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Trigonometry

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1

Contents

S.No. Topic Name Page No.

1. Measurement of Angles 1 – 7

2. Trigometric Functions (T - Ratios) 8 – 21

3. Ratios of standard angles 22 – 33

4. Conversional T – Ratios 34 – 45

5. Basic Identities 46 – 61

6. Sum and difference of angles of T-Ratios 62 - 72

7. Transformation of a Product into a sum or Difference 73 – 86

8. Multiples and sub multiples 87 – 100

9. Special Properties 101 – 115

10. Graphs of Trigonometric Ratios 116 – 125

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1

‘Radians’ and ‘Degrees’ are generally units of

measurement

Why 𝜋 = 180ᵒ:-

Consider a circle of radius r >0, circumference of the

circle c = 2𝜋r where 𝜋 = 22

7

90ᵒ o A B o A o A

B B B

B

180ᵒ 360ᵒ

(a) = 90ᵒ (b) = 180ᵒ (c) = 360ᵒ

Arc AB = 2

4

r arc AB =

2

2

r Arc AB = 2𝜋r

= 2

r = 𝜋r

Here central angle of 360ᵒ cuts off an arc of length 2𝜋r,

which is same as the circumference of the circle. So we

can say

360ᵒ equals 2𝜋r(or 2𝜋 ‘radii’)

Here r is variable and 2𝜋 = constant, so instead of using

the ‘radii’, we use the term radians;

360ᵒ = 2𝜋 radians.

⇒ 1ᵒ = 2

360

radians ⇒ 1ᵒ =

180

radians

♦ Degrees to radians – multiply by 180

x degrees = 180

x

radians

♦ Radians to degree – multiply by 180

x radians = 180

x

radians

or Radians

Degrees 180

=

♦ commonly used angles in degrees

♦ commonly used angles in radians.

♦ Arc length: -

If is the angle subtended at the center of a circle of

radius ‘r’ by an arc of length ‘l’ then

l

r

l

r=

Here l and r are in the same unit and is always in

radians.

Measurement of Angles 1

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2

1. Convert 18ᵒ to radians

Sol. 18ᵒ = 18×180

=

10

or

10

c

2. Convert 9

radians to degrees

Sol. 9

=

9

×

180

= 20ᵒ

3. Convert 11ᵒ15’ to radians

Sol. 1ᵒ = 60 minutes

15’ = 15

60degrees =

1

4degrees

11ᵒ 15’ = 1

114

+

degrees =

45

4degrees

45

4

= 45

4 180 16

= rad

4. Convert 8

to degrees

Sol. 8

=

180

8 =

180

8

= 45

2 = 22.5ᵒ = 22ᵒ

1

2×60’ = 22ᵒ30’

5. Express 45ᵒ20’10’’ in radian measure (𝜋 = 3.14)

Sol. 10’’ = 10

60min. =

10

60 60degrees =

1

360degrees

20’ = 20

60degrees =

1

3degrees

∴ 45ᵒ20’10’’ = 1 1

453 360

+ +

degrees

= 16321

360 degrees

Now

16321 16321

360 360 180

=

rad

= 16321 3.14

360 180 = 0.79 radian

6. Express 1.4 radians in degrees

Sol. 1.4 radian = 1.4×180

degrees

= 1.4×180

22×7 = (80-18)ᵒ = 80ᵒ (.18×60)’

= 80ᵒ (10.8)’ = 80ᵒ10’ (.8×60)’’ = 80ᵒ10’ 48’’

7. Find the length of an arc of a circle of radius 5cm

subtending a central angle measuring 15ᵒ.

Sol. Let l be the length of the arc

Here r = 5cm, = 15ᵒ = 15×180

rad =

12

rad

L = r = 5×5

12 12

= cm

8. Find the length of the arc of a circle of radius 3 inches

subtended by the central angle of 220ᵒ

Sol. let length of arc be l inches

Here radius = 3 inches

= 220ᵒ = 220×11

180 9

= radian

L = r = 3×11 11

9 3

= inches.

9. Find in degrees the angle subtended at the center of the

circle of diameter 30 cm by an arc of length 11cm.

Sol. length of arc = 11cm

Radius = 30

2 = 15 cm

L= r

⇒ = 11

15

l

r= radian

= 11 180

15 degrees

= 11 180

715 22

degrees = 42ᵒ

Examples



3

1. Value of 165ᵒ in radians?

(a) 11

12

(b)

33

35

(c)

5

6

(d)

7

6

2. Degree value of 11

4

radian is

(a) 415ᵒ (b) 390ᵒ (c) 435ᵒ (d) 495ᵒ

3. Convert 49ᵒ to radian

(a) ( )0.73c

(b) ( )0.65c

(c) ( )0.79c

(d) ( )0.85c

4. Which of the following is true?

(a) 415ᵒ = 11

5

(b) 235ᵒ =

43

11

(c) 270 ᵒ = 7

4

(d) 660ᵒ =

11

3

5. Value of 1ᵒ in radian?

(a) 0.1476 radian (b) 0.01746 radian

(c) 0.2836 radian (d) 0.0134

6. Value of 1 radian in degree

=

22

7

(a) 57ᵒ51’22’’ (approx.) (b) 57ᵒ21’16’’ (approx.)

(c) 57ᵒ16’22’’ (approx.) (d) 57ᵒ62,16’’ (approx.)

7. Which of the following is wrong?

(a) 4 2

5 3

+ = 264ᵒ (b)

12 15

+ = 27ᵒ

(c) 7

12 3

− = 45ᵒ (d) None of these

8. Two angles of a triangle are 35ᵒ and 45ᵒ. Third angle is

(a) 5

8

Radian (b)

5

9

radian

(c) 3

7

radian (d)

2

7

radian

9. In a right angled triangle, difference of the two non right

angles is 6

then radian value of both angles is

(a) 2

,

3

(b) ,

4 5

(c) ,

4 6

(d) ,

3 6

10. Two angles of a triangle are 30ᵒ45’15’’ and 29ᵒ14’45’’.

Third angle is

(a) 2

(b)

3

10

(c)

2

3

(d)

12

11. The circular measure of an angle of an isosceles triangle

is5

9

. Find the circular measure of one of the other

angles is -

(a) 5

18

(b)

5

9

(c)

2

9

(d)

4

9

12. Radian value of 63ᵒ14’51’’ is

(a) 2811

8000

radian (b) 3811

8000

radian

(c) 4811

800

radian (d)5811

8000

radian

13. In circular measure, the 30ᵒ30’ is equal to

(a) 51

180

radian (b)

61

180 radian

(c) 61

360 radian (d)

12

14. Radian value of 20ᵒ20’20’’ is

(a) 3601

4320

radian (b)

3661

32400 radian

(c) 2361

3240

radian (d) None of these

15. A central angle in a circle of radius 5cm cuts off an arc of

length 7.5𝜋 cm. What is the measure of angle?

(a) 150ᵒ (b) 270ᵒ (c) 135ᵒ (d) 210ᵒ

16. Find the length of the arc of a circle of radius 10 cm

subtended by central angle of 220ᵒ

Exercise - 1


4

(a) 110

9

cm (b)

90

11

cm

(c) 22

9

cm (d) 22𝜋 cm

17. Value of 45ᵒ15’ in radian is

(a) 91

180 radian (b)

91

90 radian

(c) 181

720 radian (d)

181

360 radian

18. The circular measure of the magnitude of the angle of a

regular hexagon

(a) 3

(b)

3

4

(c)

2

3

(d)

5

4

19. if three angles of a quadrilateral are 57ᵒ, 83ᵒ and 115ᵒ

then the fourth angle will be

(a) 3

4

(b)

7

12

(c)

2

3

(d)

4

3

20. If the arc of same length in two circles subtend angles of

60ᵒ and 75ᵒ at their centers. Find the ratio of their radii.

(a) 5 : 4 (b) 4 : 5 (c) 5 : 6 (d) 6 : 5

21. Difference between two angles 5

9

and 54ᵒ35’

(a) 45ᵒ35’ (b) 56ᵒ35’ (c) 46ᵒ25’ (d) 45ᵒ25’

22. Convert 0.6 radians to degrees?

(a) 38ᵒ17’22’’ (b) 35ᵒ16’22’’

(c) 34ᵒ21’49’’ (d) 48ᵒ19’47’’

23. What is the radian measure of 1

8 of a full rotation?

(a) 90ᵒ (b) 45ᵒ (c) 22.5ᵒ (d) 67.5ᵒ

24. Two angles of a triangle are 1

2 radian and

1

3 radian.

The measure of the third angle in degree (taking p =

22/7)

(a) 1

13211

(b) 2

13211

(c) 3

13211

(d) 132°

25. By decreasing 15° of each angle of a triangle, the ratios

of their angles are 2 : 3 : 5. The radian measure of

greatest angle is:

(a) 11

24

(b)

12

(c)

24

(d)

5

24



5

Answer Key 1 a 2 d 3 d 4 d 5 b

6 c 7 d 8 b 9 d 10 c

11 c 12 a 13 c 14 b 15 b

16 a 17 c 18 c 19 b 20 a

21 d 22 c 23 b 24 c 25 a

1. (a)

165ᵒ = 165×180

=

11

12

2. (d)

11

4

radian =

11 180

4

= 495ᵒ

3. (d)

49ᵒ = 49×180

radian =

49 22

180 7

=

77

90 = 0.85 rad

4. (d)

415ᵒ = 415×180

=

9

4

radian

235ᵒ = 235×47

180 36

= radian

270ᵒ = 270×180

=

3

2

radian

660ᵒ = 660×11

180 3

= radian

5. (b)

1ᵒ = 180

radian =

22

7 180 radian

= 11

630 radian = 0.01746 radian

6. (c)

1 radian = 180

degree =

1807

22 degrees

= 630

11

= 573

11

= 57ᵒ

'3 60

11

= 57ᵒ16’

''4 60

11= 57ᵒ16’22’’ (approx.)

7. (d)

4 2

5 3

+ = 144ᵒ+120ᵒ = 264ᵒ

12 15

+ = 15ᵒ + 12ᵒ = 27ᵒ

7

12 3

− = 105ᵒ - 60ᵒ = 45ᵒ

8. (b)

Sum of two angles = 35ᵒ+45ᵒ = 80ᵒ

Third angle = 180ᵒ - 80ᵒ

= 100ᵒ = 100ᵒ×180

radian =

5

9

radian

9. (d)

Triangle – right angled

∴ Sum of other two angles = 2

Difference between other two angels = 6

Let other angles be a and b

Here a + b =2

, a – b =

6

∴ a =3

, b =

6

Sum of other two angles should be 2

that is in only

option (d)

10. (c)

Sum of two angles = 30ᵒ45’15’’+29ᵒ14’45’’

= (30+29)ᵒ(45+14)’(15+45)’’ = 59ᵒ59’60’’

= 59ᵒ60’ = 60ᵒ = 3

Third angle = 𝜋 - 2

3 3

=

Solutions


6

11. (c)

In isosceles triangle – value of two angles is same.

If both angle are 5

9

then

Sum of two angles = 10

9

> 𝜋 (not possible)

Hence let two equal angels be x radian each

∴ 2x+5

9

= ⇒ x =

2

9

12. (a)

51’’ = 51' 51

60 60 60=

degree =

17

1200 degrees

14’ = 14

60 degrees =

7

30 degrees

63ᵒ14’51’’ =

+ +

7 1763

30 1200 degrees

= 75600 280 17

1200

+ +

= 75897

1200

Now,

75897

1200 180

radian =

2811

8000

radian

13. (c)

30ᵒ 30’ = 30

3060

+

degrees =

130

2

+

degrees

= 61

2 degrees

Now,61

2

=

=61 61

2 180 360 radian

14. (b)

20’’ =

= =

'20 20 1

60 60 60 180

20’ = 20 1

60 3

=

20ᵒ20’20’’ =

+ +

1 120

3 180

=

+ +

=

3600 60 1 3661

180 180

3661

180=

3661

180 180 =

3661

32400 radian

15. (b)

Radius = 5 cm

Length of arc = 7.5𝜋 cm

L = r

⇒ = 7.5

5

l

r

= =

3

2

radian =

3 180

2

= 270ᵒ

16. (a)

Let length of the arc be l cm

l = r

here r = 10 cm, = 220ᵒ = 220×180

=

11

9

L = 10×11

9

=

110

9

17. (c)

45ᵒ15’ = 15

4560

+

=181

4

181 181 181

4 4 180 720

= = radian

18. (c)

Each angle of the regular hexagon = ( )2 180n

n

−

= ( )6 2 180

6

− = 120ᵒ

120 ᵒ = 120×2

180 3

=

19. (b)

Sum of angles of quadrilateral = 360ᵒ

Sum of three angles = 57ᵒ+83ᵒ115ᵒ = 255ᵒ

Fourth angle = 360ᵒ-255ᵒ = 105ᵒ

105ᵒ = 105×7

180 12

= radian

20. (a)

Let r1 and r2 be the radii of given circles,

Length of arcs = same for both = l

For first circle = 60ᵒ = 3

∴ l = r1

r1 =

=3l l

and r2 = 12

5 5

12

l l

=

r1 : r2 = 3 12

:5

l l

= 5 : 4



7

21. (d)

5

9

radian =

5 180

9

= 100ᵒ

Now

100ᵒ - 54ᵒ35’ = 45ᵒ25’

22. (c)

0.6 radians = 0.6×180

degrees

= 0.6 180 7 378

22 11

= =

'4 4 60

34 3411 11

=

= 34ᵒ

' ''9 9 60

21 34 21'11 11

=

= 34ᵒ21’49’’

23. (b)

For full rotation = 360ᵒ

Radian measure = 2𝜋

∴ 1

8 of a full rotation =

2

8

=

4

= 45ᵒ

24. (c)

Sum of two angles = 1 1 5

2 3 6+ = radian

Third angle = 5

6

−

radian

= 22 5 132 35 97

7 6 42 42

−− = = rad

radians = 180°

97

42 radians =

180 97

42

= 180 7 97 1455 3

13222 42 11 11

= =

25. (a)

2x + 3x + 5x = 180° – 45° = 135°

10x = 135° x = 135 27

10 2=

Largest angle

= 5x + 15° = 27

52

+ 15° =

135 30 165

2 2

+ =

180° = radian

165 165 11

2 180 2 24

= = radian


8

Base (A)

In a right angle triangle

[The side opposite to angle is called perpendicular and

adjacent to angle is called base]

sin = Perpendicular

hypotenuse =

h

p,

L

K

cos = Hypotenuse

Base =

b

h,

A

K

tan = Perpendicular

Base =

p

b,

L

A

cot = 1

tan =

Perpendicular

Base =

b

p,

A

L

sec = 1

cos=

hypotenuse

base =

h

b,

K

A

cosec = 1

sin=

hypotenuse

Perpendicular =

p

h,

K

L

sin = 1

cosec ⇒ sin cosec = 1

cos = 1

sec ⇒ cos sec = 1

tan = 1

cot ⇒ tan cot = 1

sin

cos=

ph

bh

= p

b = tan

= =

cos

sin

bbh

p ph

= cot

Triplets to remember

3 4 5

6 8 10

9 12 15

5 12 13

10 24 26

7 24 25

9 40 41

11 60 61

Trigometric Functions (T-ratios) 2



9

1. Find value of sin , cos tan

12

5

Sol. Here perpendicular (L) = 5

Base (A) = 12

Hypotenuse (K) = 13

sin = =5

13

L

K

cos = =12

13

A

K

tan = =5

12

L

A

2. If sin = 11

15, then find tan

Sol.

11

sin =11

15,

sin =L

K

∴ L = 11, K = 15, A = ?

By Pythagoras theorem

K2 = L2+A2 ⇒ A2 = K2 – L2

⇒ A2 = (15)2 – (11)2 ⇒ A2 = 225 – 121

⇒ A2 = 104 ⇒ A = =104 2 26

Now tan = =11

2 26

L

A

3. If cot = 6

8, then find other trigonometric ratios

Sol.

10

8

6

cot = 6

8 =

A

L

∴ A = 6, L = 8

Now K2 = A2+L2

= 62 + 82 = 36 + 64 = 100

K = 100 = 10

sin = =8

10

L

K, cosec =

=

1 10

sin 8

cos = =6

10

A

K, sec =

=

1 10

cos 6

tan = =8

6

L

A

4. If tan = a

b, find the value of

−

+

sin cos

sin s

a b

a bco

Sol. tan = a

b =

L

A

+2 2a b

a

b

K = +2 2a b

sin =+2 2

L a

K a b

cos = =+2 2

A b

K a b

Examples


10

Now,

−

+

sin cos

sin cos

a b

a b

=

− + +

+ + +

2 2 2 2

2 2 2 2

a ba b

a b a ba b

a ba b a b

=

−

+

+

+

2 2

2 2

2 2

2 2

a b

a b

a b

a b

= −

+

2 2

2 2

a b

a b

2nd Method

tan = a

b

Now,

−

+

sin cos

sin cos

a b

a b

=

−

+

sin cos

cos cossin cos

cos cos

a b

a b

=

−

+

tan

tan

a b

a b =

−

+

aa b

ba

a bb

= −

+

2 2

2 2

a b

a b

5. If sin = a

bthen the value of sec - cos is (where 0ᵒ< <

90ᵒ)

Sol. sin = a

b =

L

K

b

a

−2 2b a

Base (A) = −2 2b a

cos = −

=2 2A b a

K b

sec =

1

cos=

−2 2

b

b a

sec - cos = −2 2

b

b a

−

−2 2b a

b

= ( )− −

−

2 2 2

2 2

b b a

b b a =

−

2

2 2

a

b b a

6. if

+

−

tan cot

tan cot = 2 then the value of sin =

Sol.

+

−

tan cot

tan cot = 2

Using componendo dividendo

( )( )

+ + −

+ − −

tan cot tan cot

tan cot tan cot =

+

−

2 1

2 1 ⇒

2tan

2cot = 3

⇒

tan

1cot

= 3 ⇒ tan2 = 3 ⇒ tan = =3

1

L

A

2

3 a

1

K = +3 1 = 2

sin = L

K =

3

2

7. For the right angle ∆ABC, Find the values of all six

trigonometric ratios of acute angles A and B.

5

3

4

B

A C

Sol. For angle A, Base = 4, perpendicular = 3, hypotenuse = 5

sin = =Perpendicular 3

hypotenuse 5

cos A = 4

=Hypotenuse 5

Base

tan = Perpendicular 3

=Base 4

cot A = 1

tan A=

4

3

sec A = 1

cos A=

5

4

cosec A = 1

sin A =

5

3

for angle B, Base (A) = 3, Perpendicular (L) = 4,

hypotenuse (K) = 5

sin B = 4

5, cos B =

3,

5

tanB = 4

3, cot B =

3

4

sec B = 5

3, cosec B =

5

4



11

1. Value of cot is

√5

1

2

(a) 1

2 (b)

5

2 (c) 2 (d)

1

5

2. Value of tan A is

√5

5

13

12

(a) 5

12 (b)

2

5 (c)

5

13 (d)

12

13

3. Values of sin , tan are

41

40

9

(a) 9 41

,40 9

(b) 40 40

,41 9

(c) 9 40

,41 410

(d) 40 41

,9 40

4. Find values of cosA, cosecA –

41

A C

15

17

8

(a) 8 17

,17 15

(b) 8 15

,17 17

(c) 17 8

,8 15

(d) 17 15

,15 8

5. Find values of tanA, cotB

11

60

A

B C

(a) 11

60,

11

60 (b)

11

60,60

11 (c)

60 61,

61 60 (d)

60

11,

60

11

6. Find the value of sinB

3 A

6

C

B D

(a) 2

3 (b)

2

13 (c)

4

5 (d)

3

13

7. In ∆ABC, ∠B = 90ᵒ c = 12 cm and a = 9cm then cos C is

(a) 3

5 (b)

3

4 (c)

3

4 (d)

4

5

8. In ∆ABC, ∠B = 90ᵒ and AB : BC = 2 :1. Then value of

(cosA + tanC) is

(a) 2 + 5 (b) ( )2 1 5

5

+ (c)

2 5

5

+ (d) 1 + 5

9. if sin = 2 2

2 2

a b

a b

−

+ then find cot

(a) 2 2

2 2

a b

a b

+

− (b)

2 2

2ab

a b+ (c)

2 2

2ab

a b− (d)

2 2

3ab

a b+

Exercise - 1


12

10. If cot = 40

find values of cosec and sec

(a) 41

40,

41

9 (b)

41

9,41

40 (c)

40

41

40

9 (d)

40

9,40

41

11. If tan = 4

3, then the value of

3sin 2cos

3sin 2cos

+

−is

(a) 0.5 (b) -0.5 (c) 3.0 (d) -3.0

12. If 5tan = 4, then 5sin 3cos

5sin 2cos

−

+ is equal to

(a) 2

3 (b)

1

4 (c)

1

6 (d)

1

3

13. If sec tan 5

sec tan 3

+=

− then sin?

(a) 1

4 (b)

1

3 (c)

2

3 (d)

3

4

14. sin cos

sin cos

+

− = 3 then the value of sin2 is

(a) 4

5 (b)

2

5 (c)

1

5 (d)

3

5

15. if tan = 1

11 and 0ᵒ<<

2

then the value of

2 2

2 2

cos sec

cos sec

ec

ec

−

+ is

(a) 3

4 (b)

4

5 (c)

5

6 (d)

6

7

16. sin cos

sin cos

+

−=

5

4then the value of

2

2

1 tan

2 tan

+=?

(a) 25

16 (b)

41

81 (c)

41

80 (d)

40

41

17. if secA = 5

4 then the value of

cot 1

cot 1

A

A

+

− is

(a) 1

4 (b)

2

7 (c)

3

7 (d) 7

18. if sin = 2 2

a

a b+ 0ᵒ < < 90ᵒ, Find the value of cos

and tan

(a) 2 2

,b a

ba b+ (b)

2 2,

+

b b

aa b

(c) 2 2

2 2,

+

+

b a b

aa b (d)

2 2

2 2,

+

+

a b a

a a b

19. if cosA = 5

13, then

cos

cos cos

ecA

A ecA+ ?

(a) 169

229 (b)

229

169 (c) 169 (d) 229

20. If cos = 3

5 , then the value of sin cos cot ?

(a) 16

25 (b)

9

16 (c)

9

25 (d)

4

3

21. If sec tan 51

2sec tan 79

+=

− then the value of the sec - tan

(a) 91

144 (b)

39

72 (c)

65

144 (d)

35

72

22. If sin = a

b, find sec + tan in terms of a and b

(a) b a

b a

+

− (b)

a b

a b

+

− (c)

b a

b a

+

− (d)

b a

a b

+

−

23. If cos = 3

5, Find the value of

sin cot

2

cot

−

(a) 160

3 (b) 2 (c)

3

160 (d) 3

24. if tan = 20

21, then

1 sin cos

1 sin cos

− +

+ +=?

(a) 3

7 (b)

7

3 (c)

20

7 (d)

7

20

25. if is an acute angle such that cos = 3

5 then

2

sin tan 1

2 tan 1

−

−

(a) 16

625 (b)

1

36 (c)

3

115 (d)

160

3



13

1. If angle C of triangle ABC is 90ᵒ, then tanA + tanB =?

(Where a,b,c are sides opposite to angles A, B, C

respectively)

(a) ab

c (b)

2 2

2

a b

c

− (c)

2c

ab (d)

2

2 2

c

a b−

2. tanA = 2 1− then cosA= ?

(a) 4 2 2− (b) 1

4 2 2−

(c) 4 (d) 1

4 2−

3. Which of the following is true for a right angled

triangle?

(a) sin2 + cos2 = 1 (b) sec2 - tan2 = 1

(c) cosec2 - cot2 = 1 (d) All of these

4. In a right angled triangle XYZ right angled at Y, XY = 2√6

cm and XZ – YZ = 2, then sec X + tan X is

(a) 1

6 (b) 6 (c) 2 6 (d)

6

2

5. If cosA = 2 2

3, then cosA cosecA + tanA secA = ?

(a) 16 2 3

8

+ (b)

7 5 3

8

+

(c) 16 2 5

7

+ (d)

16 7 5

7

+

6. In ∆PQR, right angled at Q, PR+QR = 25cm and PQ = 5

cm determine the values of sinP, cosP and tanP.

(a) 13 5 5

, ,12 13 12

(b) 12 5 12

, ,13 13 5

(c) 13 13 12

, ,12 5 5

(d) 12 13 13

, ,13 5 12

7. If 3cos = 1 find the value of 2 26sin tan

4cos

+

(a) 10 (b) 13 (c) 0 (d) 9

8. If secA = 5

4 then

3

3

3sin 4sin

4cos 3cos

−

−

A A

A A

(a) 44

117 (b)

117

44

− (c)

217

22 (d)

22

217

−

9. If is an acute angle and tan2 = 8

7 then the value of

( )( )

( )( )

1 sin 1 sin

1 cos 1 cos

+ −

+ −

(a) 7

8 (b)

8

7 (c)

7

4 (d)

64

49

10. In triangle ABC, right angled at C and AB = 29cm, BC =

21cm, ∠ABC = then find out value of cos2 + sin2

(a) 41

841 (b)

3

2 (c)

801

841 (d) 1

11. In a right angled triangle ABC, right angled at B, tan A =

1 then value of 2sinA cos A=?

(a) 1

2 2 (b) 1

(c) 2 2 (d) none of these

Exercise - 2


14

Answer key

1 c 2 b 3 b 4 a 5 d

6 b 7 a 8 b 9 c 10 a

11 c 12 c 13 a 14 a 15 c

16 b 17 d 18 a 19 a 20 c

21 c 22 a 23 c 24 a 25 c

1. (c)

Here L = 1, A = 2, K = √5

cot = ( )( )

Base

perpendicular

A

L=

2

1 = 2

2. (b)

Here

L = 12, A = 5, K = 13

tanA = Perpendicular(L)

Base(A)=

12

5

[Note:- the side opposite to angle is called

perpendicular and adjacent to angle is called base]

3. (b)

Here

L = 40, A = 9, K = 41

sin = =40

41

L

K

tan = =40

9

L

A

4. (a)

Here

L = 15, A = 8, K = 17

cosA = =8

17

A

K

cosecA = =17

15

K

L

5. (d)

For angle A

L = 60, A = 11, K = 61

∴tanA = L

A =

0

11

For angle B

L = 11, A = 60, k = 61

cot B = A

L =

60

11

6. (b)

∆BDC is right angle triangle

For angle B

A = 6, L = 4

K = 2 26 4+ = 52 = 2 13

sinB = 4 2

2 13 13

L

K= =

7. (a)

b

12

9

A

C B

(Note - c is the side opposite to angle C and a and b are

also sides opposite to angles ∠A and ∠B respectively.)

b = 2 212 9+ = 15

here L = 12 cm A = 9cm, K = 15 cm

cos C = 9 3

15 5

A

K= =

8. (b)

Let AB and BC be 2 and 1 respectively.

Then AC = 2 22 1 5+ =

Solutions

Exercise - 1



15

5

1

2

C

A B

For angle A,

L = 1, A = 2, K = 5

cos A = 2

5

A

K=

For angle C,

L =2, A = 1 K = 5

tanC = 2

1 = 2

cosA + tanC = 2

5+ 2 =

( )2 1 5

5

+

9. (c)

2 2a b+

2 2a b−

A

C B Here

L = 2 2a b− , k = 2 2a b+

A = 2 2−K L = ( ) ( )2 2

2 2 2 2+ − −a b a b

= ( )4 4 2 2 4 4 2 22 2+ + − + −a b a b a b a b = 2ab

∴ cot = 2 2

2A ab

L a b=

−

10. (a)

41 40

9

A

C B

cot = 9

40

Here L = 40, A = 9

9, 40 and 41 are triplet

∴ K = 41

Now cosec = 41

40

k

l=

sec 41

9

k

A=

11. (c)

tan = 4

3

now 3sin 2cos

3sin 2cos

+

− =

sin cos3 2

cos cossin cos

3 2cos cos

+

−

= 3tan 2

3tan 2

+

− =

43 2

34

3 23

+

−

= 6

2 = 3

12. (c)

5 tan = 4 ⇒ tan = 4

5

5sin 3cos

5sin 2cos

−

+ =

5 tan 3

5 tan 2

−

+ =

45 3

54

5 25

−

+

= 1

6

13. (a)

sec tan 5

sec tan 3

+=

−

Using C & D

⇒ ( )

( )

sec tan sec tan 5 3

sec tan sec tan 5 3

+ + + +=

− − − − ⇒

2sec

2 tan

= 4

⇒

1

cossin

cos

⇒ 1

sin = 4 ⇒ sin =

1

4

Alternate:

sec tan 5

sec tan 3

+=

−

3(sec + tan) = 5(sec-tan) ⇒ 3sec + 3tan = 5sec-5tan

⇒ 2sec = 8 tan ⇒ 2 sin

8cos cos

= ⇒ sin =

2 1

8 4=

14. (a)

sin cos

sin cos

+

− = 3 ⇒ sin +cos = 3 sin - 3cos

⇒ 2sin = 4cos ⇒ sin 4

cos 2

= = 2 ⇒ tan = 2


16

2

1

A

C B

here L = 2, A = 1, K = 5

sin = 2

5

L

K=

sin2 =

2

2 4

55

=

15. (c)

1

A

C B

tan = 1

11

L = 1, A = 11

K = 11 1+ = 12 = 2 3

cosec = 2 3

1 = 2 3

sec = 2 3

11

w 2 2

2 2

cos sec

cos sec

ec

ece

−

+

=

( )

( )

22

22

2 32 3

11

2 32 3

11

−

+

=

1212

120 51112 144 6

1211

−

= =

+

16. (b)

sin cos

sin cos

+

− =

5

4

⇒ 4(sin + cos) = 5(sin − cos)

⇒ sin = cos ⇒ tan = 9

Now

2

2

1 tan 1 81

2 812 tan

+ +=

=

82 41

2 81 81=

17. (d)

5

3

4

secA = 5

4 =

h

b

L = 2 25 4− = 3

∴ cot A = 4

3

cotA 1

cot 1A

+

−=

41

34

13

+

−

=

7

31

3

= 7

18. (a)

2 2a b+

a

b

Here L = a

K = 2 2a b+

A = 2 2 2a b a+ − = b

cos = 2 2

A b

K a b=

+

tan = L a

K b=

19. (a)

13

12

5

Base (A) 2 213 5− = 12

∴ cosec A = 13

12

K

L=



17

cos

cos cos

ecA

A ecA+ =

13

125 13

13 12+

=

13

1260 169

156

+ =

13 156 169

229 12 229

=

20. (c)

sin cos cot = sin cos cos

sin

= cos2 =

23 9

5 25

=

21. (c)

sec tan

sec tan

+

−=

512

79 ⇒

sec tan 209

sec tan 79

+=

−

Using C & D

⇒ ( )( )

sec tan sec tan

sec tan sec tan

+ + −

+ − − =

209 79

209 79

+

−

⇒ 2sec 288

2 tan 130

= ⇒

1

144cossin 65

cos

= ⇒ sin = 65

144

22. (a)

b

a

2 2b a−

sin = a

b

Base (A) = 2 2b a−

sec = 2 2

=−

K b

A b a

tan = 2 2

L a

A b a=

−

sec + tan = 2 2 2 2

+− −

b a

b a b a

= 2 2

b a

b a

+

− =

+

+ −

b a

b a b a=

b a

b a

+

−=

b a

b a

+

−

23. (c)

5

4

3

cos = 3

5, sin =

4

5

cot = 3

4

Now,

4 3

sin cot 5 42 2

3cot

4

−−

= =

1

208

3

= 3

160

24. (a)

29

20

21

L = 20, A = 21

sin = 20

29 cos =

21

29

1 sin cos

1 sin cos

− +

+ + =

20 211

29 2920 21

129 29

− +

+ +

=

29 20 21

2929 20 21

29

− +

+ + =

30 3

70 7=

25. (c)

5

4

3

cos = 3

5

∴ sin = 4

5, tan =

4

3

Now

2

sin tan 1

2 tan 1

−

− =

4 41

5 316

2 19

−

−

=

1

1523

9

= 9 3

15 23 115=


18



19

Answer key

1 c 2 b 3 d 4 b 5 a

6 b 7 a 8 b 9 a 10 d

11 b

1. (c)

c

b

a

A

B C

tanA = a

b

tanB = b

a

tanA + tanB = a b

b a+ =

2 2a b

ab

+ =

2c

ab

2. (b)

1

B

A C

tan = 2 1−

K = ( )2

22 1 1− + = 2 1 2 2 1+ − + = 4 2 2−

cosA = 1

4 2 2=

−

base

hypotenuse

3. (d)

Let ABC a right angled triangle right angled at B

A

B c

Here sin = AB

AC, cosec =

AC

AB

cos = BC

AC, sec =

AC

BC

tan = AB

BC, cot =

BC

AB

and AC2 = AB2 + BC2

sin2 + cos2 = 2 2

AB BC

AC AC

+

= 2 2 2

2 2

AB BC AC

AC AC

+= = 1 (true)

sec2 - tan2 = 2

2

AC

BC-

2

2

AB

BC =

2 2

2

AC AB

BC

−

= 2

2

BC

BC = 1 (true)

cosec2 - cot2 = 2

2

AC

AB-

2

2

BC

AB=

2 2

2

AC BC

AB

−

= 2

2

AB

AB = 1 (true)

4. (b)

2√6

Z

X Y

Let YZ be a cm then XZ = a + 2

Then in ∆ XYZ

(XZ)2 = (YZ)2 + (XY)2

⇒ (a+2)2 = a2 + (2√6)2

⇒ a2+4a+4 = a2+24 ⇒ 4a = 20 ⇒ a = 5

∴ YZ = 5cm, XZ = 7 cm

secX +tanX = 7 5

2 6 2 6+ =

126

2 6=

Exercise - 2


20

5. (a)

3

1

2 2

B

A C

cosA = 2 2

3

BC = ( )2

23 2 2− = 1

cosecA = 3, secA = 1 3

cos 2 2A=

tanA = 1

2 2

cosA cosec A + tanA secA

= 2 2 1 3

33 2 2 2 2

+ = 2√2 +3

8 =

16 2 3

8

+

6. (b)

5

R

P Q

Let QR = x cm

Then PR = 25 – x cm

In ∆PQR

PR2 = PQ2+QR2

(25-x)2 = 52+x2 ⇒ 625+x2-50x = 25+x2

⇒ 50x = 600 ⇒ x = 12cm

∴ QR = 12cm, PR = 13cm

∴sinP = 12

13

cos P = 5

13

tan P = 12

5

7. (a)

3

2√2

1

cos = 1

3

sin = 2 2

3, tan = 2 2

2 26sin tan

4cos

+=

86 8

91

43

+

= 120 3

9 4

= 10

8. (b)

5

3

4

C

A B

BC = 2 25 4− = 3

sin A = 3

5

cos A = 1 4

sec 5A=

Now

3

3

3sin 4sin

4cos 3cos

A A

A A

−

− =

3

3

3 33 4

5 5

4 44 3

5 5

−

−

=

9 108

5 125256 12

125 5

−

−

= 225 108

256 300

−

−= -

117

44

9. (a)

tan2 = 8

7

tan = 2 2

7



21

15 2√2

7

( )( )

( )( )

1 sin 1 sin

1 cos 1 cos

+ −

+ −

= 2

2

1 sin

1 cos

−

− [∵ (a+b)(a-b) = a2 – b2]

Here = sin = 2 2

15

L

K=

cos = 7

15

A

K=

=

2

2

2 21

15

71

15

−

+

=

81

157

115

−

−

=

7

7158 8

15

=

10. (d)

21

A

B C

AC = 2 2 2 229 21AB BC− = −

= ( )( )29 21 29 21− + = 8 50 400 = = 20 cm

Now

cos = 21

29

A

K=

sin = L

K =

20

29

cos2+sin2 = 2 2

21 20

29 29

+

= 441 400

841

+ = 1

Note : Value of sin2 + cos2 will be always 1

irrespective of the value of

11. (b)

C

A B

tan A = 1 = BC

AB

∴ BC = AB = 1

and AC 2 2BC AB= + = 1 1 2+ =

sin A = 1

2

C

AC

=

cos A = 1

2

AB

AC=

2sin A cos A = 2×1 1

2 2 = 1


22

0ᵒ 30ᵒ 45ᵒ 60ᵒ 90ᵒ

sin 0 1

2

1

2 3

2 1

cos 1 3

2

1

2

1

2 0

tan 0 1

3 1 3

Not

define

cot Not

define 3 1

1

3 0

sec 1 2

3 2 2

Not

define

cosec Not

define 2 2

2

3 0

Examples 1. Find the value of tan260ᵒ cosec30ᵒ tan45ᵒ

Sol. tan260ᵒ cosece30ᵒtan45ᵒ

= ( )2

3 ×2×1 = 6

2. Find the value of (cos0ᵒ + sin45ᵒ + sin30ᵒ)(sin90ᵒ-

cos45ᵒ+cos60ᵒ)

Sol. (cos0ᵒ+sin45ᵒ+sin30ᵒ)(sin90ᵒ-cos45ᵒ+cos60ᵒ)

= 1 1 1 1

1 12 22 2

+ + − +

= 2 2 1 2 2 1

2 2

+ + − +

= 3 2 3 2

2 2

+ −

= 9 2 7

4 4

−=

3. Value of cosec30ᵒ + cot45ᵒ?

Sol. cosec30ᵒ + cot45ᵒ

= 2 + 1 = 3

4. If = 45ᵒ then 2sin cos =?

Sol. = 2sin45ᵒ cos45ᵒ

= 2×1 1

2 2 = 1

5. sin30 tan 45 sin60 cos30

sin 45 sec60 cot 45 sin90

+ − −

Sol.

1 3 3

12 2 21 2 1 1

2

+ − −

= 1 1 3 3

2 2 22+ − −

= 1 1

322

+ − = 2 1 2 3

2

+ −

6. Value of 2 2 25sin 30 cos 45 4 tan 30

2sin30 cos30 tan 45

+ −

+

Sol. 2 2 25sin 30 cos 45 4 tan 30

2sin30 cos30 tan 45

+ −

+

=

2221 1 1

5 42 2 3

1 32 1

2 2

+ −

+

=

1 1 15 4

4 2 3

31

2

+ −

+

=

5 1 4

4 2 3

3 1

2

+ −

+ =

15 6 16

12

3 1

2

+ −

+

= 5 2

12 3 1

+ =

( )5

6 3 1+

7. If 3tan22x = cos60ᵒ+sin45ᵒcos45ᵒ then x =?

Sol. 3tan22x = cos60ᵒ+sin45ᵒcos45ᵒ

3tan22x = 1 1 1

2 2 2+

Ratios of Standard Angles 3



23

⇒ 3tan22x = 1 1

2 2+ = 1

⇒ tan22x = 1

3 ⇒ tan2x =

1

3= tan30ᵒ

⇒ 2x = 30ᵒ ⇒ x = 15ᵒ

8. If sin(x-y) =1

2 and cos(x+y)=

1

2 then value of x is

Sol. sin(x-y) = 1

2

⇒ sin(x-y) = sin30ᵒ

⇒ (x-y) = 30ᵒ ...................... (i)

cos(x+y) = 1

2

⇒ cos (x+y) = cos (60ᵒ)

⇒ (x+y) = 60ᵒ .................. (ii)

(i) + (ii)

2x = 90ᵒ

X = 45ᵒ

Y = 60 - 45ᵒ = 15ᵒ

9. In ∆ABC, right angled at B, AB = 3cm and AC = 6cm find

∠BAC and ∠ACB

Sol.

6

3

A

C B

BC = 2 26 3 3 3− =

sinC = 3 1

6 2

AB

AC= = = sin30ᵒ

∴ ∠ACB = 30ᵒ

∠BAC = 90 – 30ᵒ = 60ᵒ


24

1. cos60ᵒ cos30ᵒ - sin60ᵒsin30ᵒ = ?

(a) 1 (b) 0 (c) 1

2 (d) -1

2. tan230ᵒ+tan260ᵒ+tan245ᵒ

(a) 13

3 (b)

1

12 (c)

13

2 (d)

1

13

3. cos30 sin60

1 cos60 sin30

+

+ + =?

(a) 3 (b) 3

4 (c)

3

2 (d) 1

4. 4cot260ᵒ + sec230ᵒ-sin245ᵒ =?

(a) 5

2 (b)

17

6 (c)

13

6 (d)

19

6

5. 2

2 tan30

1 tan 30

+ =?

(a) sin60ᵒ (b) cos60ᵒ (c) tan60ᵒ (d) sin30ᵒ

6. cos60ᵒcos45ᵒ + sin60ᵒsin45ᵒ

(a) 2

3 1+ (b)

3 1

3 1

−

+ (c)

3 1

2 2

− (d)

2 2

3 1+

7. sin230ᵒ+sin245ᵒ+sin260ᵒ+sin290ᵒ

(a) 2

5 (b)

3

5 (c)

5

2 (d)

5

3

8. if 3 sinx = cosx then x = ?

(a) 30ᵒ (b) 45ᵒ (c) 60ᵒ (d) 0ᵒ

9. Find value of cot22 2 2cot cot cot

2 3 4 6

+ + +

(a) 15

4 (b)

17

4 (c)

13

3 (d)

14

3

10. cosec290ᵒ + cosce2 60ᵒ + cosece2 45ᵒ + cosce2 30ᵒ

(a) 23

4 (b)

17

6 (c)

13

3 (d)

25

3

11. tan60ᵒ cosece245ᵒ + sec260ᵒtan245ᵒ

(a) 4+ 2 3 (b) 4 - 2 3 (c) 2 - 3 (d) 2 + 3

12. 2sin230ᵒtan260ᵒ - 3cos260ᵒsec230ᵒ

(a) 1

2 (b)

2

3 (c)

3

4 (d)

4

3

13. tan 45

cos 30ec

+

sec60 5sin90

cot 45 2cos0

−

=?

(a) 0 (b) 2

5 (c) 1 (d)

1

5

14. 4 4 2 2 22 cos sin tan cot 3sec

3 6 3 4 6

+ − + +

=?

(a) 1

4 (b)

2

3 (c)

4

7 (d)

4

3

15. sin230ᵒcos245ᵒ + 4tan230ᵒ +1

2sin290ᵒ - 2cos290ᵒ +

1

24

cos0ᵒ

(a) 1 (b) -1 (c) 0 (d) 2

16. 2 2 2

2 2

cos 60 4sec 30 tan 45

sin 30 cos 30

+ −

+

(a) 64

3 (b)

55

12 (c)

67

12 (d)

67

10

17. (cosec245ᵒsec230ᵒ)(sin230ᵒ+4cot345ᵒ-sec260ᵒ)

(a) 1

2 (b) (c)

2

3 (d)

3

2

18. tan30ᵒsec45ᵒ + tan60ᵒsec30ᵒ

(a) 2 2 3

3

+ (b)

3 2 2

3

+

(c) 2 2 3

2 3

− (d)

3 2 2

2 3

−

Exercise - 1



25

19. sin30 cos45

tan 60

=?

(a) 6

12 (b)

2

3 (c) 2 3 (d)

2

3

20. If sinx = 3 cosx then cosec2

x =?

(a) 2

3 (b) 2 (c) 2 (d) Not define

21. 2sin2

x = 1 then cosx =?

(a) 3

2 (b)

1

2 (c)

3

2 (d)

1

2

22. 2cos2

2

x = 1 than tanx =?

(a) 3 (b) 1

3 (c) 1 (d) not defined

23.

2 2 2 2

2 2 2 2

cos 45 cos 60 tan 30 sin 30

sin 60 sin 45 cot 45 cot 30

+ − −

(a) 1

4 (b)

3

4 (c)

1

2 (d)

5

4

24. sin3

3

cot

6

- 2sec2

4

+3 cos

3

tan

4

- tan2

3

(a) 35

8 (b)

35

8− (c)

17

4− (d)

11

8


26

1. if sin(A-B) = sinA cosB – cosA sinB then find value of

sin15ᵒ

(a) 3 2

2

− (b)

3 1

2 2

+ (c)

3 2

2 2

− (d)

3 1

2 2

−

2. if 2sin2x+cos245ᵒ = tan45ᵒ & 0ᵒ < x < 90ᵒ then sec2x = ?

(a) 1

2 (b)

2

3 (c) 2 (d) 2

3. 2 2

2 2

cos 30 sec 45

8cos 45 sin 60

x ec

= tan2 60ᵒ - tan2 30ᵒ, then x =?

(a) 0 (b) 1 (c) 2 (d) 2

4. x sin60ᵒtan30ᵒ - tan245ᵒ = cosec60ᵒcot30ᵒ - sec245ᵒ

then x = ?

(a) 2 (b) -2 (c) 6 (d) -4

5.

2

2 2

2

2 tan6 sec sec 0 sec

4 31 tan

6

x

+ + =

−

then x = ?

(a) 2 (b) 1 (c) 0 (d) -1

6. if x cos45ᵒ = y sec60ᵒthen 4

4

x

y

(a) 43 (b) 63 (c) 23 (d) 83

7. If tan (A+B) = 3 and tan (A-B) =1

3; 0ᵒ < A+B ≤ 90ᵒ;

A>B then find A and B?

(a) ,3 4

(b) ,

3 6

(c) ,

2 6

(d) ,

4 12

8. If and are positive acute angles sin (4−) = 1 and

cos (2+) =1

2, then the value of cos (+) is

(a) 0 (b) 1 (c) 3

2 (d)

1

2

9. If sin + cos = 2 (0ᵒ ≤ , ≤ 90ᵒ) then tan2

3

+

= ?

(a) 1

3 (b) 3 (c) 1 (d) 1

10. In an acute angled triangle ABC, if tan (A+B-C) = 1, sec

(B+C-A) = 2 and sin (C+A-B) = 1

2, find the value of A, B

and C

(a) 1 1

77 , 45 , 222 2

A B C

= = =

(b) 1 1

52 , 55 ,C 372 2

A B

= = =

(c) 1 1

60 , 52 ,C 672 2

A B

= = =

(d) 1 1

37 , 52 , C 452 2

= = =A B

11. If tan (+) = 3 and sec (1-2) = 2

3 then the value of

sin21 + tan32 is equal to (0 < − < + ᵒ)

(a) 0 (b) 3 (c) 2 (d) 1

12. If tanA = 1, tan B = 3 , then 1 1

1 cos1 sin

2

++

−BA

=?

(a) 2 1

2 3

− (b)

3 2 2

2 1

+

+ (c)

3 1

2 2

− (d)

2 3

2 1+

13. If 2

2 2

cos

cot cos

− = 3 and 0ᵒ < < 90ᵒ then the value of

is : -

(a) 30ᵒ (b) 45ᵒ (c) 60ᵒ (d) None of these

14. If sin(A+B)= 1

2 and sin (A-B) =

1

2 then (cos2 B – cos2 A

) = ?

(a) 1

2 (b) 1 (c) 0 (d) 2

15. Which of the following is true (0ᵒ < < 90ᵒ)

(a) if 2 cos 3 = 1 then = 20ᵒ

(b) if 2 sin 2 = 3 then = 9ᵒ

Exercise - 2



27

(c) if tan 5 = 1 then = 30ᵒ

(d) All of these

16. Find an acute angle , when sin cos 3 1

sin cos 3 1

− −=

+ +

(a) 60ᵒ (b) 90ᵒ (c) 45ᵒ (d) 30ᵒ

17 If sin2x = sin60ᵒ cos30ᵒ - cos60ᵒ sin30ᵒ then find the

value of x?

(a) 55ᵒ (b) 100ᵒ (c) 65ᵒ (d) 15ᵒ

18. If each of is a positive acute angle such that cosec

(+−) = 2

3 sec (+−) = 2 and cot (+−) = 1, find

the value of , ?

(a) 1 1

33 , 45 , 522 2

= = =

(b) 1

82 , 602

= = =

(c) 1

52 , 602

= = =

(d) 1

52 , 602

= = =

19. if cosec = 2 then 2 2

2 2

2sin 3cot

4 tan cos

+

−

(a) 4

5 (b)

3

4 (c)

8

7 (d)

1

7

20. If + = 90ᵒ, and = 2 then cos2 + sin2 + tan2 +

cot2 =

(a) 15 (b) 15

4 (c)

13

3 (d)

17

4

21. if 2

2

tan 30

1 cot 60

x x−

+ = sin230ᵒ+4cot245ᵒ- cosec230ᵒ the

value of x2 is:

(a) 1

4 (b)

1

3 (c)

1

2 (d)

1

5

22. if = 60ᵒ then 1

1 sin 1 sin

+

+ − = ?

(a) 8 (b) 0 (c) 4 (d) 2

23. If cosecA = 2 find the value of 1 sin

tan 1 cos

A

A A+

+

(a) 1 (b) 0 (c) -1 (d) 2

24. Find acute angle A and B. if sin (A+2B) = 3

2 and cos

(A+4B) = 0, A > B

(a) A = 30ᵒ, B = 15ᵒ (b) A = 30ᵒ, B = 45ᵒ

(c) A = 60ᵒ, B = 15ᵒ (d) A = 55ᵒ, B = 30ᵒ

25. If + =2

and cos =

3

2, then value of sec =?

(a) 1 (b) 2

3 (c) 2 (d) 2


28

Answer key 1 b 2 a 3 c 4 c 5 c

6 a 7 c 8 a 9 c 10 d

11 a 12 a 13 a 14 a 15 d

16 b 17 c 18 a 19 a 20 c

21 b 22 d 23 b 24 b 25 c

1. (b)

cos60ᵒ cos30ᵒ - sin60ᵒsin30ᵒ

= 1

2×

3 3 1

2 2 2− = 0

2. (a)

tan230ᵒ+tan260ᵒ+tan245ᵒ

= ( )2

213 1

3

+ +

=

1

3+3+1 =

13

3

3. (c)

cos30 sin60

1 cos60 sin30

+

+ + =

3 3

32 21 1 2

12 2

+

=

+ +

4. (c)

4cot260ᵒ + sec230ᵒ-sin245ᵒ

= 4

2 2 2

1 2 1

3 3 2

+ −

= 4 4 1 8 1 16 3 13

3 3 2 3 2 6 6

−+ − = − = =

5. (a)

2

2 tan30

1 tan 30

+ =

2

12

3

11

3

+

=

2

3

11

3+

=

2

3

4

3

= 2 3

4 3

=

3

2 = sin60ᵒ

6. (c)

cos60ᵒcos45ᵒ + sin60ᵒsin45ᵒ

= 1 1 3 1

2 22 2 + =

3 1

2 2

+

7. (c)

sin230ᵒ+sin245ᵒ+sin260ᵒ+sin290ᵒ

=

222

21 1 31

2 22

+ + +

= 1 1 3

14 2 4

+ + + = 1 5

1 12 2

+ + =

8. (a)

3 sinx = cosx ⇒ sin 1

cos 3

x

x= ⇒ tanx =

1

3

tanx = tan30ᵒ

X = 30ᵒ

9. (c)

cot22 2 2cot cot cot

2 3 4 6

+ + +

= ( ) ( )2

2210 1 3

3

+ + +

= 1 13

1 33 3

+ + =

10. (d)

cosec290ᵒ + cosce2 60ᵒ + cosece2 45ᵒ + cosce2 30ᵒ

= ( ) ( )2

2 22 21 2 2

3

+ + +

= 1+

4

3+2+4 =

25

3

11. (a)

tan60ᵒ cosece245ᵒ + sec260ᵒtan245ᵒ

= ( ) ( )2 2

3 2 2 1+ = 2 3 4+

12. (a)

Solutions

Exercise - 1



29

2sin230ᵒtan260ᵒ - 3cos260ᵒsec230ᵒ

( )2 2

21 1 22 3 3

2 2 3

−

= 1 1 4

2 3 34 4 3

− = 3 1

12 2

− =

13. (a)

tan 45

cos 30ec

+

sec60 5sin90

cot 45 2cos0

−

= 1 2 5

2 1 2+ − =

5 50

2 2− =

14. (a)

4 4 2 2 22 cos sin tan cot 3sec

3 6 3 4 6

+ − + +

= ( ) ( )24 4

2 21 1 22 3 1 3

2 2 3

+ − + +

= 1 1 4

2 3 1 316 16 3

+ − + +

=

1 14 4

4 4− + =

15. (d)

sin230ᵒ cos245ᵒ + 4tan230ᵒ + 1

2sin290ᵒ - 2cos290ᵒ +

1

24

cos0ᵒ

= ( ) ( ) ( )22

1 1 1 1 14 1 2 0 1

2 2 242 3

+ + − +

=1 1 1 1 1

4 04 2 3 2 24

+ + − +

= 1 4 1 1

8 3 2 24+ + +

3 32 12 1

24

+ + += = 2

16. (b)

2 2 2

2 2

cos 60 4sec 30 tan 45

sin 30 cos 30

+ −

+

=

22

22

1 24 1

2 3

1 3

2 2

+ −

+

=

1 161

3 64 12 554 3

1 12 12

+ −+ −

= =

17. (c)

(cosec245ᵒsec230ᵒ)(sin230ᵒ+4cot345ᵒ-sec260ᵒ)

= ( ) ( ) ( )2 2

2 3 22 12 4 1 2

23

+ −

= 8 1 2

4 43 4 3

+ − =

18. (a)

tan30ᵒsec45ᵒ + tan60ᵒsec30ᵒ

= 1 2

2 33 3

+ = 2 2 2 3

23 3

++ =

19. (a)

sin30 cos45

tan 60

=

1 1

12 2

3 2 6

= = 6

12

20. (c)

sinx = 3 cosx ⇒ sin

3cos

x

x= ⇒ tanx = 3 ⇒ x = 60ᵒ

Now cosec 60

2

= cosec30ᵒ = 2

21. (b)

2sin2

x = 1 ⇒ sin

2

x =

1

2

⇒ sin2

x = sin30ᵒ ⇒

2

x = 30ᵒ, x = 60ᵒ

∴ cos60ᵒ = 1

2

22. (d)

2cos2

2

x = 1 ⇒ cos2

1

2 2

x=

⇒ cos2

x =

1

2 = cos 45ᵒ ⇒

2

x = 45ᵒ

X = 90ᵒ

Now tan90ᵒ = not defined

23. (b) 2 2 2 2

2 2 2 2

cos 45 cos 60 tan 30 sin 30

sin 60 sin 45 cot 45 cot 30

+ − −

= ( ) ( )

22 2 2

2 2 2 2

11 1 1

2 232

113 3

22

+ − −

= 1 1 2 1 1

2 3 4 1 3 4 3

+ − −

= 2 1 1 1

3 2 3 12+ − − =

8 6 4 1 9 3

12 12 4

+ − −= =


30

24. (b)

( ) ( )3

2 23 13 2 2 3 1 3

2 2

− + −

= 3 3 3

3 2 2 38 2

− + − = 9 3

4 38 2

− + −

= 9 3

78 2

+ − = 9 12 56 35

8 8

+ −= −

25. (c)

cos =3

2⇒ = 0ᵒ ⇒ = − = ᵒ

sec = sec60ᵒ = 2

Answer key 1 d 2 c 3 b 4 a 5 a

6 a 7 d 8 d 9 b 10 d

11 c 12 b 13 c 14 a 15 d

16 a 17 d 18 c 19 c 20 c

21 a 22 a 23 d 24 a 25 c

1. (d)

15ᵒ = 45ᵒ - 30ᵒ

Sin (45ᵒ-30ᵒ) = sin45ᵒcos30ᵒ-cos45ᵒsin30ᵒ

= 1 3 1 1

2 22 2− =

3 1

2 2

−

2. (c)

2sin2x +

2

1

2

= 1 ⇒ 2sin2x + 1

2 = 1

⇒ 2sin2x = 1

2 ⇒ sin2x =

1

4 ⇒ sinx =

1

2

x = 300 (0ᵒ < x < 90ᵒ)

sec 2x = sec 60ᵒ = 2

3. (b)

2 2

2 2

cos 30 sec 45

8cos 45 sin 60

x ec

= tan2 60ᵒ - tan2 30ᵒ

⇒( ) ( )

( )22 2

2

22

2 2 13

31 38

22

= −

x

⇒ 4 2 13

1 3 38

2 4

x = −

⇒ 8 8

3 3

x= ⇒ x = 1

4. (a)

3 1

2 3x - 1 = ( )

223 2

3 −

⇒ 12

x− = 2 – 2 = 0 ⇒

2

x = 1, x = 2

5. (a)

2

2 2

2

2 tan6 sec sec 0 sec

4 31 tan

6

x

+ + =

−

Exercise - 2



31

⇒

12

31

13

−

+2+1 = 2x ⇒ 1+3 = 2x ⇒ x = 2

6. (a)

x cos45ᵒ = y sec60ᵒ

x ×1

2 = y×2 ⇒

x

y = 2 2

⇒ ( ) ( )4

4 442 2 2 2

x

y

= =

= 24×22= 43

7. (d)

tan (A+B) = 3 = tan60ᵒ

(A+B) = 60ᵒ =3

....... (i)

tan (A-B) = 1

3 = tan30ᵒ

(A-B) = 30ᵒ =6

...... (ii)

(i) + (ii)

2A = 3 6 2

+ = ⇒ A =

4

And B = 3 4 12

− =

8. (d)

sin (4−) = 1 = sin90ᵒ

4− = 90ᵒ ....... (i)

cos (2+) = 1

2 = cos60ᵒ

2+ = 60ᵒ ….. (ii)

(i) + (ii)

= 150ᵒ ⇒ = 25ᵒ

And = 60ᵒ - 2×25ᵒ = 10ᵒ

Now

cos (+) = cos (25ᵒ+20ᵒ) = cos 45ᵒ = 1

2

9. (b)

sin+ cos = 2 possible iff sin = cos = 1

∴ = 90ᵒ, = 0ᵒ

Now tan 2

3

+

= tan2 90 0

3

+

= tan60ᵒ = 3

10. (d)

tan (A+B-C)=1 = tan45ᵒ

⇒ A+B-C = 45ᵒ ......... (i)

sec (B+C-A) = 2 = sec60ᵒ

⇒ B+C-A = 60ᵒ ........ (ii)

sin(C+A-B) = 1

2 = sin 30ᵒ

⇒ C+A-B = 30 ....... (iii)

(i)+(ii) 2B = 105ᵒ ⇒ B = 521

2

(ii)+(iii) 2C = 90ᵒ ⇒ C = 45ᵒ

(i)+(iii) 2A = 75ᵒ ⇒ A = 371

2

11. (c)

tan (+) = 3 = tan 60ᵒ

⇒ + = ᵒ (i)

sec (1−)= 2

3

⇒ − = 30ᵒ ....... (ii)

(i)+(ii), 2 = 90ᵒ

⇒ = 45ᵒ

And = 60ᵒ - 45ᵒ = 15ᵒ

Now

sin 21 + tan 3 = sin 90ᵒ+tan45ᵒ = 1+1 = 2

12. (b)

tanA = 1 A = 45ᵒ

tanB = 3 , B = 60ᵒ

1 1 1 1

1 cos 1 cos 45 1 sin 301 sin

2

BA+ = +

+ + − −

= 1 1

1 11 1

22

+

+ −

= 2 2 2 2 2

22 1 2 1

+ ++ =

+ +=

3 2 2

2 1

+

+

13. (c)

2

2 2

cos

cot cos

− = 3

⇒ cos2 = 3cot2 - 3cos2 ⇒ 4cos2 = 3 cot2

⇒ 4 cos2 = 32

2

cos

sin

⇒ sin2 =

3

4

⇒ sin = 3

2 ⇒ = 60ᵒ

14. (a)


32

sin(A+B) = 1

2 = sin 45ᵒ

⇒ (A+B) = 45ᵒ .............. (i)

sin (A-B) = 1

2 = sin 45ᵒ

⇒ A – B = 45ᵒ ……(ii)

(i) + (ii)

2A = 90ᵒ ⇒ A = 45ᵒ

and B = 0

now cos2B – cos2A = cos20ᵒ - cos245ᵒ

= 1 -

2

1 1 11

2 22

= − =

15. (d)

(a) cos3 = 1

2 = cos60ᵒ

⇒ 3 = 60ᵒ, = 20ᵒ (true)

(b) sin2 = 3

2 = sin60ᵒ

⇒ 2 = 60ᵒ, = 30ᵒ (true)

(c) tan5 = 1 = tan 45ᵒ

⇒ 5 = 45ᵒ, = 9ᵒ (true)

16. (a)

sin cos 3 1

sin cos 3 1

− −=

+ +

By using C & D

( ) ( )( ) ( )

( )( ) ( )

3 1 3 1sin cos sin cos

sin cos sin cos 3 1 3 1

− + +− + +=

− − + + − −

⇒ 2sin 2 3

2cos 2

= ⇒ tan = 3 = tan 60ᵒ ⇒ = 60ᵒ

Alternate:

sin cos

sin cos

−

+ =

3 1

3 1

−

+

=

3 1

2 2

3 1

2 2

−

+

= sin 60 cos60

sin 60 cos60

−

+

∴ = 60ᵒ

17. (d)

sin2x = sin60ᵒ cos30ᵒ - cos60ᵒsin30ᵒ

sin2x = 3 3 1 1

2 2 2 2 − =

3 1

4 4−

⇒ sin2x = 1

2 ⇒ 2x = 30ᵒ

x = 15ᵒ

18. (c)

cosec (+−) = 2

3

+− = 60ᵒ ....... (i)

sec (+−)

+− = 60ᵒ ….... (ii)

cot (+−) = 1

⇒ +− = 45ᵒ ........ (iii)

(i) +(ii) 2 = 120ᵒ ⇒ = 60ᵒ

(ii)+(iii) 2 = 105ᵒ = =521

2

(i) +(iii) 2 = 105ᵒ ⇒ = 521

2

19. (c)

cosec = 2 = cosec45ᵒ

⇒ = 45ᵒ

Now 2 2

2 2

2sin 3cot

4 tan cos

+

−

=

( )

( )

2

2

2

2

12 3 1

2

14 1

2

+

−

=

12 3

21

42

+

−

= 4

7

2

= 8

7

20. (c)

+ = 90ᵒ , = ⇒ = 90ᵒ, = 30ᵒ

∴ = 90ᵒ - 30ᵒ = 60ᵒ

Now cos2+sin2+tan2+cot2

= cos2ᵒ+sin2ᵒ+tan2ᵒ+cot2ᵒ

= ( )2 22

21 3 13

2 2 3

+ + +

= 1 3 1

34 4 3

+ + + = 1 13

43 3

+ =

21. (a)

2

2

tan 30

1 cot 60

x x−

+ = sin230ᵒ+4cot245ᵒ - cosec230ᵒ

2

2

1

3

11

3

x x

−

+

= ( ) ( )2

2 214 1 2

2

+ −



33

⇒ 13 4 4

1 41

3

xx −

= + −

+

⇒

2

134 4

3

x

= ⇒ 1

2 4

x= ⇒ x =

1

2

∴ x2 = 2

1 1

2 4

=

22. (a)

1

1 sin 1 sin

+

+ − =

1

1 sin60 1 sin60

+

+ −

= 1 1

3 31 1

2 2

+

+ −

= 2 2

2 3 2 3+

+ −

= ( )( )4 2 3 4 2 3

2 3 2 3

− + +

+ − =

8

1 = 8

23. (d)

cosecA = 2, ∴ A = 30ᵒ

=

1

1 sin 30 1 21tan 30 1 cos30 3

13 2

+ = +

+ +

= 3 + 1

2 3+ =

2 3 3 1

2 3

+ +

+ =

4 2 3

2 3

+

+ = 2

24. (a)

sin(A+2B) = 3

2 ⇒ (A+2B) = 60ᵒ ............ (i)

and cos (A+4B) ⇒ (A+4B) = 90ᵒ ........... (ii)

(ii)-(i)

2B = 30ᵒ ⇒ B = 15ᵒ

Now, A+2×150 = 60ᵒ ⇒ A = 30ᵒ

25. (c)

+ = 2

, cos =

3

2, = 30ᵒ =

6

Now = 2

-

6

=

3

⇒ ∴ sec = sec

3

= 2

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Trigonometry

Conversion T-Rations

Chapter - 04


34

2

3

2

0, 2𝜋 𝜋

2nd quadrant

sin

cosec (+ve)

1st quadrant

All (+ve)

cos

sec (+ve)

4th quadrant

tan

cot (+ve)

3rd quadrant

♦ Trigonometric ratios change for the odd multiples of 90ᵒ

Ex. 90ᵒ, 270ᵒ, 450ᵒ, 630ᵒ ................ etc

sin ↔ cos

tan↔ cot

sec ↔ cosec

♦ Trigonometric ratios do not change for the even

multiples of 90ᵒ

Ex. 180ᵒ, 360ᵒ, 540ᵒ, 720ᵒ etc

90ᵒ

270ᵒ

0, 360ᵒ

180ᵒ

2nd quadrant

90+

180 -

1st quadrant

90 -

360+

-

270+

360-

4th quadrant

180+

270-

3rd quadrant

1st quadrant

90 - is in 1st quadrant and ratios will change.

In first quadrant all are positive

sin (90 - ) = cos

cos (90 - ) = sin

tan(90 - ) = cot

cot (90 - ) = tan

sec (90 - ) = cosec

cosec (90 - ) = sec

2nd quadrant

( sin, cosec → positive )

for (90+), ratios change

sin (90 + ) = cos

cos (90 + ) = - sin

tan(90+) = - cot

cot (90+ ) = - tan

sec (90 + ) = - cosec

cosec (90 + ) = sec

for (180-), ratios don’t change

sin (180 - ) = sin

cos (180 - ) = -cos

tan(180 - ) = - tan

cot (180 - ) = -cot

sec (180 - ) = - sec

cosec (180 - ) = cosec

3rd quadrant

( tan, cot → positive )

For (180+), ratios don’t change

sin (180+) = - sin

cos (180+ ) = -cos

tan(180+) = tan

cot (180+) = cot

sec (180+) = - sec

cosec (180+) = - cosec

For {(270-)→ odd multiple of 90ᵒ}, ratios change

sin (270 - ) = -cos

cos (270 - ) = -sin

tan(270 - ) = cot

cot (270 - ) = tan

sec (270 - ) = - cosec

cosec (270 - ) = - sec

Conversion of T - Ratios 4



35

4th quadrant

(sec, cos → positive)

For (270+) → ratios change

sin (270 + ) = - cos

cos (270 + ) = sin

tan(270 + ) = - cot

cot (270 + ) = -tan

sec (270 + ) = cosec

cosec (27 + ) = - sec

(360-) ≌ (-)

For (-), (360-) ratios don’t change

sin ( - ) = -sin

cos (- ) = cos

tan(- ) = - tan

cot (- ) = - cot

sec (- ) = sec

cosec (- ) = - cosec

Ex. Solve

(i) sec (540ᵒ - ) (ii) tan7

2

+

(iii) cos (11𝜋 - ) (iv) sin 9

2

+

Sol. (i) sec (540-)

[ 540ᵒ = 360ᵒ + 180ᵒ∴ (540ᵒ-) is in 2nd quadrant]

= sec (6×90ᵒ - ) = - sec

(ii) tan 7

2

+

⇒

74

2 2

+ = − +

∴ 7

2

+

is in 4th quadrant] = tan 7

2

+

= - cot

(iii) cos(11𝜋-)

[ 11𝜋 - = (5×2𝜋) + (𝜋-) 2nd quadrant] = - cos

(iv) sin 9

2

+

[

9

2

+ = 2×2𝜋 +

2

+

,

2

+ is in 2nd quadrant]

= cos

Ex. Find value: -

(i) sin 120ᵒ (ii) cos150ᵒ

(iii) tan 225ᵒ (iv) sec3

4

(v) cosec 7

4

(vi) cot

9

4

Sol. (i) sin120ᵒ = sin(90ᵒ+30ᵒ) = cos (30ᵒ)= 3

2

(ii) cos150ᵒ = cos (180-30ᵒ) = -cos(30)= 3

2−

(iii) tan225ᵒ = tan(180ᵒ+45ᵒ) = tan45ᵒ = 1

(iv) sec3

4

= sec

4

−

= - sec

4

= 2−

(v) cosec7

4

= cosec 2

4

−

= - cosce

4

= 2−

(vi) cot9

4

= cot 2

4

+

= cot

4

= 1

Ex. Find value:

(i) sin (750ᵒ) (ii) cos (960ᵒ)

(iii) tan 675ᵒ (iv) cot21

4

(v) sec53

4

Sol. (i) sin(750ᵒ) = sin (2×360ᵒ+30ᵒ) = sin 30ᵒ = 1

2

(ii) cos (960ᵒ) = cos (5×180ᵒ+60ᵒ) = - cos60ᵒ = -1

2

(iii) tan 675ᵒ = tan(2×360ᵒ-45ᵒ) = - tan 45ᵒ = -1

(iv) cot21

4

= cot 5

4

+

= cot

4

= 1

(v) sec53

4

= sec 13

4

+

= - sec

4

= 2−

Ex. If A+B+C = 2

then value of sin (A+B) is

(a) sinC (b) cosC (c) – sinC (d) – cosC

Sol. (b)

A+B+C =2

⇒ A+B =

2

- C

⇒ sin(A+B) = sin2

C

−

= cosC

Trigonometry ratios of complementary angles

If sin x = cos y then x + y = 900

sin x = cos y = sin (900 - y)

x = 900 – y x + y = 900

Thus

If tan x = tan y then x + y = 900

and sec x = sec y then x + y = 900


36

Ex. ( )cos37

sin 90 37

− =?

Sol. ( )cos37

sin 90 37

− =

cos37

cos37

= 1

Ex. cosec25ᵒ - sec65ᵒ =?

Sol. cosec25ᵒ - sec65ᵒ = cosec25ᵒ - sec (90-25ᵒ)

= cosec25ᵒ - cosec25ᵒ = 0

Ex. cos80

sin10

+ cos59ᵒcosec31ᵒ =?

Sol. cos80

sin10

+ cos59ᵒcosec31ᵒ

= ( )

cos80

sin 90 80

− +cos59ᵒcosec(90-59ᵒ)

= cos80

cos80

+cos59ᵒsec59ᵒ = 1+

cos59

cos59

= 1+1 = 2

Ex. sin2 21° + sin2 69° is equal to:

Sol. sin2 21° + sin2 69° = sin2 21° + sin2 (900 - 21°)

= sin2 21° + cos2 21° = 1

Ex. sin2 1° + sin2 2° + …… + sin2 89° + sin2 90° = ?

Sol. Let the number of terms be n, then

n = last term first term

common difference

−+ 1

n = 90 1

1

−+ 1 =90

sin21° + sin2 2° + … + sin2 45° + … + sin2 89° + sin2 90°

= (sin2 1° + sin2 89°) + (sin2 2° + … + sin2 88°) + … + to

44 terms + sin2 45°+ sin2 900

= (sin2 1° + cos2 1°) + (sin2 2° + cos220) + … + to 44

terms + sin2 45°+ sin2 900 = 44 + 1

2+ 1

= 2 21

45 sin cos 12

+ =



37

1. value of cos(675ᵒ) is

(a) 1

2 (b)

1

2− (c)

1

2 (d)

3

2

2. Value of tan 1650ᵒ is

(a) 3

2 (b)

1

3 (c)

1

2− (d)

1

3

3. cos(-660ᵒ) = ?

(a) 3

2 (b)

1

2 (c)

1

2− (d)

1

2+

4. which of the following is true?

(a) sec 480ᵒ = 2 (b) sin 135ᵒ = 1

2

(c) cos 210ᵒ = 3

2− (d) All of these


(a) tan (180ᵒ– ) = - tan (b) sin (360 - ) = - sin

(c) sec (270ᵒ - ) = - cosec (d) cos (270+) = - sin

6. cot 3

2

−

= ?

(a) tan (b) - tan (c) cot (d) - cot

7. Value of tan225ᵒcot405ᵒ + tan765ᵒcot675ᵒ is

(a) 1 (b) 0 (c) -1 (d) -2


(a) sin (-65ᵒ) = -cos25ᵒ

(b) cosec (-225ᵒ) = - cosec45ᵒ

(c) tan (135ᵒ) = - tan45ᵒ

(d) cos 18ᵒ = sin72ᵒ


(a) tan (𝜋 - ) = - tan (b) cot (-) = - cot

(c) cos (-) = cos (d) sec (-) = - sec

10. The line that makes 640ᵒ angle, lies in

(a) 1st quadrant (b) 2nd quadrant

(c) 3rd quadrant (d) 4th quadrant

11. If cos = 1

2 then value of is:

(a) 60ᵒ, 120ᵒ (b) -60ᵒ, -120ᵒ

(c) 60ᵒ, 300ᵒ (d) 120ᵒ, -300ᵒ

12. If A+B+C = 2

then value of sec (B+C)

(a) cosA (b) secA (c) cosecA (d) - cosecA

13. If A+B+C = 𝜋 then value of tan 2

A B+

is

(a) tan C (b) tan2

C (c) - cot

2

C (d) cot

2

C

14. Which of the following is right

(a) cosec2 (90ᵒ - ) – tan2 = 1

(b) sec2 (90ᵒ - ) – cot2 = 1

(c) sin2 (90ᵒ - ) + sin2 = 1

(d) All of these

15. sinA cos (90ᵒ+A) + cosA sin(270ᵒ+A) = ?

(a) 1 (b) - 1 (c) 0 (d) 2

16. sin 18ᵒ + sin 54ᵒ + sin 150ᵒ + sin 198ᵒ + sin 234ᵒ=?

(a) 1

2− (b)

3

2 (c)

1

2 (d) -

3

2

17. sin + cos = 3 sin (90ᵒ-) then cot = ?

(a) 3 1

2

+ (b)

3 1

2

− (c) 3 (d)

3

2

18. ( ) ( )

( )

sin 90 cos 90

cot 90

− −

− =?

(a) 1+sin2 (b) 1- sin2 (c) – sin2 (d) 2- sin3

19. ( ) ( ) ( )

( )

cos 90 sin sin 90 cos sin 90 sin

cos cos 90

− − −+

−

(a) 1 (b) 2 (c) -1 (d) 3

20. Evaluate tan54 cos 32 sin 41

cot 36 sec58 cos49

ec + +

Exercise - 1


38

(a) 1 (b) 0 (c) 3 (d) 3

2

21. sin 33 cos57

cos57 sin 33

+

- 4 sin245ᵒ

(a) 1 (b) 1

2 (c) -1 (d) 0

22. cos9ᵒ + cot81ᵒ = ?

(a) cos81ᵒ + tan9ᵒ (b) sin81ᵒ + tan9ᵒ

(c) sin81ᵒ + cos81ᵒ (d) sin9ᵒ+tan9ᵒ

23. Value of tan14ᵒ tan43ᵒ tan 47ᵒ tan76ᵒ is

(a) 2 (b) 3 (c) 1 (d) 4

24. The value of cot10ᵒ cot20ᵒ cot60ᵒ cot70ᵒ cot80ᵒis

(a) 1 (b) -1 (c) 3 (d) 1

3

25. if sin sec29ᵒ = 1, then value of is

(a) 29ᵒ (b) 61ᵒ (c) 31ᵒ (d) 59ᵒ



39

1. (sin 670 + cos 230) (sin 670 - cos 230)

(a) 3 (b) 0 (c) 1 (d) 2

2. If tan (7θ + 280) = cot (300 - 3θ) then value of θ is

(a) 80 (b) 50 (c) 600 (d) 90

3. If x, y are positive acute angles and x + y < 900 and sin

(2x - 200) = cos (2y + 200), then value of sec (x + y) is

(a) 2 (b) 1

2 (c) 1 (d) 0

4. sin (x + y) = cos [3(x + y)] then value of cot2(x + y) is

(a) 3 (b) 1 (c) 0 (d) 1

3

5. If (2θ + 450) and 3θ are acute angles and tan (2θ + 450)

= cot3θ, then the value of θ is:

(a) 50 (b) 90 (c) 120 (d) 150

6. If sec (4θ - 500) = cosec (500 - θ) and 00 < θ < 900 then

the value of θ is

(a) 3301

3 (b) 180 (c) 23

01

3 (d) 300

7. If tan 7θ tan 2θ = 1, then the value of tan 3θ is

(a) 3 (b) 1

3− (c)

1

3 (d) 3−

8. Value of (tan 10 tan 20 tan 30 ..... tan 890)

(a) 1 (b) 245 (c) Not defined (d) 0

9. The value of cos 10 cos 20 cos 30..... cos 1770 cos 1780 cos

1790

(a) 0 (b) 1

2 (c) 1 (d)

1

2

10. The value of cot 420 cot 430 cot 440 cot 450 cot 460 cot

470 cot 480 is

(a) 0 (b) 1 (c) - 1 (d) 3

11. ( )

( ) ( )

0 0 0

0 0 0

sin300 tan240 sec 390

cos210 cot 135 cosec 315

− − −

=?

(a) 3 (b) 2 (c) 2

3 (d)

3

2

12. If sin ∝ sec (300 + ∝) = 1 (00< ∝ < 600) then sin ∝ + cos

2∝ =?

(a) 1 (b) 2 3

2 3

+ (c) 0 (d) 2

13. If A, B and C be the angles of a triangle, then which of

the following is incorrect?

(a) sin A B

2

+

= cosC

2 (b) cos

A B

2

+

= sinC

2

(c) tan A B

2

+

= sec C

2 (d) cot

A B

2

+

= tanC

2

14. If Q is a positive acute angle and tan2θtan3θ = 1 then

the value of (2 cos25

2

-1) is

(a) -1

2 (b) 1 (c) 0 (d)

1

2

15. If A + B = 900, then

+

−2

2

tanAtanB tanAcotB sin B

sinAsecB cos A

(a) tan A (b) cot B (c) tan B (d) sin B

16. If tan150 = 2 3− then tan 150 cot 750 + tan 750 cot 150

=?

(a) 14 (b) 12 (c) 10 (d) 8

17. Find value of cot20

cot

3

20

cot

5

20

cot

7

20

cot

9

20

(a) - 1 (b) 1

2 (c) 0 (d) 1

18. tan (x + y) tan (x - y) = 1 then tan 2x

3

= ?

(a) 1

3 (b)

2

3 (c) 3 (d) 1

19. If A = tan110 tan290, B = 2 cot610cot790 then which of

Exercise - 2


40

the following is true?

(a) A = 2B (b) A = - 2B (c) 2A = B (d) 2A = - B

20. sin2 5° + sin2 6° + …… + sin2 84° + sin2 85° = ?

(a) 1

392

(b) 1

402

(c) 40 (d) 1

392

21. The value of sin2 5° + sin2 10° + sin2 15° + …… + sin2 85°

+ sin2 90° is:

(a) 1

72

(b) 1

82

(c) 1

102

(d) 1

102

22. The value of sin2 25° + sin2 45° + sin2 65° + sin2 85° is

equal to:

(a) 1.5 (b) 2 (c) 2.5 (d) 3

23. The value of sin2 1° + sin2 5° + sin2 9° + …… + sin2 89°:

(a) 1

112

(b) 11 2 (c) 11 (d) 11

2

24. If sin2 θ - 2cos θ + 1

4= 0 then θ =? (00 < θ < 900)

(a) 3

(b)

6

(c)

2

(d)

3

2



41

Answer key

1 a 2 d 3 b 4 d 5 d

6 b 7 b 8 b 9 d 10 d

11 c 12 c 13 d 14 d 15 b

16 c 17 a 18 b 19 a 20 c

21 b 22 d 23 c 24 d 25 b

1. (a)

cos (675ᵒ) = cos (2×360ᵒ-45ᵒ)

[675ᵒ= 720ᵒ-45ᵒ ∴ 4th quadrant] = cos45ᵒ = 1

2

2. (d)

tan(1650ᵒ)

[here 1650ᵒ = (360×4+210ᵒ)

∴ 1650ᵒ is in 3rd quadrant and in 3rd quadrant tan and

cot are positive]

= tan (360ᵒ×4 + 210ᵒ) = tan(210ᵒ) = tan(180ᵒ+30ᵒ)

= tan30ᵒ =1

3

3. (b)

cos(-) = cos

∴ cos(-660ᵒ) = cos660

cos(660ᵒ) = cos(2×360ᵒ - 60ᵒ) [4th quadrant]

= cos60ᵒ = 1

2

4. (d)

Option (a)

sec480ᵒ = sec(360ᵒ+120ᵒ) = sec(5×90ᵒ+30ᵒ)

= - cosec30ᵒ (2nd quadrant) = - 2 (true)

Option (b)

sin135ᵒ = sin (90+45ᵒ)

= cos45ᵒ (2nd quadrant) = 1

2 (true)

Option (c)

cos210ᵒ = cos(180ᵒ+30ᵒ)

= - cos30ᵒ (3rd quadrant) = 3

2− (true)

5. (d)

cos (270+) = sin

(270ᵒ+) is in 4th quadrant and cos is positive in 4th

quadrant

270ᵒ is odd multiple of 90ᵒ

∴ cos → sin

6. (b)

cot 3

2

−

= cot -

3

2

−

= - cot 3

2

−

[ as cot (-) = - cot ] = - tan

7. (b)

tan225ᵒ cot405ᵒ + tan 765ᵒcot675ᵒ

= tan (180ᵒ+ 45ᵒ) cot (360ᵒ+ 45ᵒ) + tan (720ᵒ+ 45ᵒ) cot

(720ᵒ- 45ᵒ) = tan45ᵒ cot45ᵒ + tan45ᵒ (-cot45ᵒ)

= tan45ᵒ cot45ᵒ - tan45ᵒ cot 45ᵒ = 0

8. (b)

Option (a)

sin (-65ᵒ) = - sin65ᵒ = - sin(90ᵒ-25ᵒ) = - cos25ᵒ

Option (b)

cosec (-225ᵒ) = - cosec 225ᵒ

= - cosec (180ᵒ+45ᵒ) = cosec45ᵒ

Option (c) tan135ᵒ = tan (180ᵒ- 45ᵒ) = - tan45ᵒ

Option (d)

cos18ᵒ = cos (90ᵒ-72ᵒ) = sin72ᵒ

9. (d)

sec and cos are positive in 4th quadrant hence sec (-) =

sec and cos(-) = cos

10. (d)

4th quadrant

Solutions

Exercise - 1


42

640ᵒ = 360ᵒ×2 - 80ᵒ

∴ 4th quadrant

11. (c)

cos = 1

2 = cos(60ᵒ) = cos(360ᵒ-60ᵒ) = cos (300ᵒ)

∴ = 60ᵒ, 300ᵒ

12. (c)

A+B+C = 2

⇒ B+C =

2

-A

⇒ sec (B+C) = sec 2

A

−

⇒ sec (B+C) = cosecA

13. (d)

A+B+C = 𝜋

A+B = 𝜋 – C ⇒ 2

A B+=

2 2

C−

⇒ tan2

A B+

= tan2 2

C −

= cot

2

C

14. (d)

Option (a)

cosec2 (90ᵒ - ) – tan2

sec2 - tan2 = 1

Option (b)

sec2(90ᵒ - ) – cot2

cosec2 - cot2 = 1

Option (c)

sin2(90ᵒ - ) + sin2 = 1

cos2 + sin2 = 1

15. (b)

sin A cos(90ᵒ+A) + cosA sin(270ᵒ+A)

= sin A (-sinA) + cos (-cosA)

= -[sin2A + cos2A] = - 1

16. (c)

sin18ᵒ + sin54ᵒ + sin150ᵒ + sin198ᵒ + sin 234ᵒ

= sin18ᵒ + sin54ᵒ + sin (180ᵒ-30ᵒ) + sin(180ᵒ+18ᵒ) + sin

(180ᵒ+54ᵒ)

= sin18ᵒ + sin54ᵒ + sin30ᵒ + (-sin18ᵒ) + (-sin54ᵒ)

= sin30ᵒ = 1

2

17. (a)

sin + cos = 3 sin (90ᵒ-)

⇒ sin + cos = 3 cos

⇒ sin = ( )3 1− cos ⇒ cos 1

sin 3 1

=

−

cot = ( )

( )( )

3 11 3 1

23 1 3 1

+ +=

− +

18. (b)

( ) ( )

( )

sin 90 cos 90

cot 90

− −

−

= cos sin

tan

=

cos sin

sin

cos

=cos2 = 1-sin2

19. (a)

( ) ( ) ( )( )

cos 90 sin sin 90 cos sin 90 sin

cos cos 90

− − −+

−

= sin sin cos cos cos sin

cos sin

+

= sin2 + cos2 =1

20. (c)

tan54 cos 32 sin 41

cot 36 sec58 cos49

ec + +

= ( ) ( ) ( )

tan54 cos 32 sin 41

cot 90 54 sec 90 32 cos 90 41

ec + +

− − −

=tan54 cos 32 sin 41

tan54 cos 32 sin 41

ec

ec

+ +

= 3

21. (d)

sin 33 cos57

cos57 sin 33

+

- 4 sin245ᵒ

sin33ᵒ = sin(90ᵒ-57ᵒ) = cos57ᵒ

∴

2

cos57 cos57 14

cos57 cos57 2

+ −

= 1 + 1 - 2 = 0

22. (b)

cos9ᵒ + cot81ᵒ

= cos(90ᵒ - 81ᵒ) + cot(90ᵒ - 9ᵒ) = sin81ᵒ + tan9ᵒ

23. (c)

tan14ᵒ tan43ᵒ tan 47ᵒ tan76ᵒ



43

= tan14ᵒ tan43ᵒ tan (90ᵒ-43ᵒ) tan(90ᵒ-14ᵒ)

= tan14ᵒ tan43ᵒ cot43ᵒ cot14ᵒ

= (tan14ᵒcot14ᵒ)(tan43ᵒcot43ᵒ) = 1

1

tancot

=

24. (d)

cot10ᵒ cot20ᵒ cot60ᵒ cot70ᵒ cot80ᵒ

= cot10ᵒ cot20ᵒ cot60ᵒ cot(90ᵒ-20ᵒ) cot(90ᵒ-10ᵒ)

= cot10ᵒ cot20ᵒ cot60ᵒ tan20ᵒ tan10ᵒ

= (cot10ᵒtan10ᵒ) (cot20ᵒtan10ᵒ) cot60ᵒ

= 1×1×1

3 =

1

3

25. (b)

sin sec29ᵒ = 1 ⇒ sinᵒ = 1

sec 29 = cos29ᵒ

⇒ sin = cos (90ᵒ- 61ᵒ) = sin61ᵒ

∴ = 61ᵒ

Answer key 1 b 2 a 3 a 4 b 5 b

6 d 7 c 8 a 9 a 10 b

11 b 12 a 13 c 14 c 15 a

16 a 17 d 18 a 19 c 20 b

21 d 22 c 23 a 24 a

1. (b)

(sin 670 + cos 230) (sin 670 - cos 230)

= [sin 670 + cos (90 - 670)] [sin670 - cos(90 - 670)]

= (sin 670 + sin 670) (sin 670 - sin 670) = 2sin670 × 0 = 0

2. (a)

tan (7θ + 280) = cot (300 - 3θ)

⇒ tan (7θ + 280) = cot [900 - (600 + 3θ)]

⇒ tan (7θ + 280) = tan (600 + 3θ)

⇒ 7θ + 280 = 600 + 3θ ⇒ 4θ = 320 ⇒ θ = 80

[If tan x = cot y then x + y = 900]

3. (a)

sin (2x - 200) = cos (2y + 200)

⇒ sin (2x - 200) = cos [900 - (700 - 2y)]

⇒ sin (2x - 200) = sin (700 - 2y)

⇒ 2x - 200 = 700 - 2y

⇒ 2x + 2y = 900 ⇒ x + y = 450

sec (x + y) = sec 450 = 2

(sin A = cos B, then A + B = 900)

here 2x - 200 + 2y + 200 = 900 ⇒ x + y = 450

4. (b)

sin (x + y) = cos [3 (x + y)]

= sin [900 - 3(x + y)] ⇒ x + y = 900 - 3(x + y)

⇒ 4(x + y) = 900 ⇒ 2(x + y) = 450

Now cot2(x + y) = cot 450 = 1

5. (b)

tan (2θ + 450) = cot 3θ

∴ (2θ + 450) + 3θ = 900 ⇒ 5θ = 450 ⇒ θ = 90

6. (d)

sec (4θ - 500) = cosec (500 - θ)

⇒ sec (4θ - 500) = cosec [90 - (400 + θ)]

⇒ sec (4θ - 500) = sec (400 + θ)

⇒ 4θ - 500 = 40 + θ ⇒ 3θ = 900 ⇒ θ = 300

(sec A = cosec B 00 < A, B < 900

then A + B = 900)

Exercise - 2


44

here 4θ - 500 + 500 - θ = 900 ⇒ 3θ = 900 ⇒ θ = 300

7. (c)

tan7θ tan 2θ = 1

⇒ tan7θ = 1

tan2= cot 2θ

⇒ tan 7θ = tan (90 - 2θ) ⇒ 7θ = 900 - 2θ

⇒ 9θ = 900 ⇒ θ = 100

Now tan 300 = 1

3

8. (a)

tan 10 tan 20 tan 30..... tan 890

= (tan10 tan890)(tan20 tan880)..... tan450

= (tan10cot10)(tan20 cot20) ....... tan 450

= 1 × 1 × ......... × 1 = 1

9. (a)

cos 10 cos 20 cos 30 ..... cos 1770 cos 1780 cos 1790

= cos 10 cos 20 cos 30..... cos 900 ....... cos 1780 cos 1790

= 0 (∵cos900 = 0)

10. (b)

cot 420 cot 430 cot 440 cot 450 cot 460 cot 470 cot 480

= cot 420 cot 430 cot 440 cot 450 tan 420 tan 430 tan 440

= (cot 420tan 420)(cot 430tan 430)(cot 440tan 440 )cot

450 = 1

11. (b)

sec (-θ) = sec θ

= ( ) ( ) ( )( )( )( )

− +

+ − −

0 0 0 0 0

0 0 0 0

sin 360 60 tan 180 60 sec 390

cos 180 30 cot135 cosec315

= ( ) ( )

( ) ( )

0 0 0 0

0 0 0 0 0

sin60 tan60 sec 360 30

cos30 cot 180 45 cosec 360 45

− +

− − −

= ( )( )

0 0 0

0 0 0

sin60 tan60 sec30

cos30 cot 45 cosec45

−

− − −=

3 23

2 3

31 2

2

= 2

12. (a)

sin ∝ = ( )0

1

sec 30 + ⇒ sin ∝ = cos (300 + ∝)

∴ ∝ + 300 + ∝ = 900 ⇒ 2∝ = 600, ∝ = 300

Now sin ∝ + cos 2∝ = sin300 + cos 600

= 1 1

2 2+ = 1

13. (c)

A + B + C = π

then A B C

2 2 2

+ = −

tan A B

2

+

= tan C

2 2

−

⇒ tan

A B

2

+

= cot C

2

14. (c)

tan 2θ tan 3θ = 1

tan2θ = cot3θ

∴ 2θ + 3θ = 900

5θ = 900

θ = 180

Now

02 5 18

2 cos 12

−

= 2 cos2450 - 1 = 2 ×

21

2

- 1 = 0

15. (a)

+−

2

2

tanAtanB tanAcotB sin B

sinAsecB cos A

= ( ) ( )( )

( )0 0 2 0

20

tanAtan 90 A tanAcot 90 A sin 90 A

cos AsinAsec 90 A

− + − −−

−

= 2 2

2

tanAcot A tan A cos A

sinAcosecA cos A

+−

= 21 tan A 1+ − = 2tan A = tanA

16. (a)

tan150 = 2 3− = tan (900-750) = cot750

cot150 = ( )( )0

1 1 2 3

tan15 2 3 2 3 2 3

+= =

− − +

= 2 3

4 3

+

−= 2 3+

cot 150 = tan 750

now tan 150 cot 750 + tan 750 cot 150

= ( )2 3− ( )2 3− + ( )2 3+ ( )2 3+

= 4 + 3 - 2 3 + 4 + 3 + 2 3 = 14

17. (d)

cot20

cot

3

20

cot

5

20

cot

7

20

cot

9

20

= cot 20

cot

3

20

cot

4

cot

3

2 20

−

cot

2 20

−

= cot20

cot

3

20

cot

4

tan

3

20

tan

20



45

=

3tan cot cot tan cot

20 20 20 20 4 = 1

18. (a)

tan (x + y) tan (x - y) = 1

tan (x + y) = ( )1

tan x y− ⇒ tan (x + y) = cot (x - y)

∴ (x + y) + (x - y) = 900 ⇒ 2x = 900,

now tan

090

3

= tan 300 =1

3

19. (c)

A = tan110tan290

B = 2cot610cot790 = 2 cot(900 - 290) cot (900-110)

= 2 tan290 tan110 = 2A

∴ B = 2A

20. (b)

Let the number of terms be n, then


common difference

−+ 1

n = 85 5

1

−+ 1 = 81

sin2 5° + sin2 6° + … + sin2 45° + … + sin2 84° + sin2 85°

= (sin2 5° + sin2 85°) + (sin2 6° + … + sin2 84°) + … + to

40 terms + sin2 45°

= (sin2 5° + cos2 5°) + (sin2 6° + … + sin2 45°) + … + to 40

terms + sin2 45°

= 40 + 2 21 1

40 sin cos 12 2

= + =

21. (d)

no. of terms = last term first term

common difference

− + 1

= 90 5

5

− + 1 = 17 + 1 = 18

(sin2 45° + sin2 85°) + (sin2 10° + sin2 80°) + … to 8

terms + sin2 45° + sin2 90°

= (sin2 5° + cos2 5°) + (sin2 10° + cos2 10°) + … to 8

terms + 1

2 + 1 = 8 +

1

2 + 1 =

19

2

[ sin (90° – ) = cos ; sin2 + cos2 = 1]

22. (c)

sin2 25° + sin2 85° + sin2 25° + sin2 65° + sin2 45°

= sin2 5° + sin2 (90° – 5°) + sin2 25° + sin2 (90° – 25°) +

sin2 45°

= sin2 5° + cos2 5° + sin2 25° + cos2 25° = sin2 45°

= 1 + 1 + 1 1

2 2.52 2

= =

23. (a)


common difference

−+ 1 ⇒ n =

89 1

4

−+ 1 = 23

sin2 1° + sin2 89° + sin2 5° + sin2 85° + … + to 22 terms +

sin2 45°

= (sin2 1° + cos2 1°) + (sin2 5° + cos2 5°) + … + to 11

terms +

2

1

2

= 2 21 1

11 11 sin cos2 2

+ = +

24. (a)

sin2 θ - 2cos θ + 1

4= 0

⇒ 1-cos2 θ - 2 cos θ + 1

4= 0 ⇒ cos2 θ + 2 cos θ -

5

4= 0

cos θ =

− + 5

2 4 4 14

2

= 2 9

2

− =

2 3 1

2 2

− = (0 < θ < 900)

∴ θ = 3


46

* * * Some Basic Identities

h

b

P

h2 = p2 + b2 .......... (i)

eq. (i) divide both sides by h2

1 = 2 2

2 2

p b

h h+

⇒ sin2 + cos2 = 1

eq.(i) divide both side by b2

2 2

h p

b b

=

+ 1

sec2 = tan2+1

⇒ sec2 - tan2 = 1

eq. (i) divide both sides by p2

2 2

1h b

p p

= +

⇒ cosec2 = 1+ cot2

⇒ cosec2 − cot2 = 1

As sec2 - tan2 = 1

From a2 – b2 = (a-b)(a+b)

(sec - tan) (sec + tan) = 1

⇒ sec + tan = 1

sec tan −

and

cosec2 - cot2 = 1

⇒ (cosec - cot) (cosec + cot) = 1

⇒ cosec + cot = 1

cos cotece −

***

If a cos - bsin = c,

and asin + bcos = k

then a2+b2 = c2+k2

Proof: a cos - b sin = c ..... (i)

a sin + b cos = k ..... (ii)

(i)2 + (ii)2

a2cos2 + b2sin2 - 2ab sin cos + a2sin2 + b2cos2 +

2ab sin cos = c2+k2

a2 (cos2 + sin2 ) + b2 (cos2 + sin2 ) = c2 + k2

⇒ a2 + b2 = c2 k2 ⇒ k = 2 2 2a b c + −

Basic Identities 5



47

Ex.1 sec2 θ + cosec2 θ =?

Sol. sec2 θ + cosec2 θ

= 2 2

2 2 2 2

1 1 sin cos

cos sin cos sin

+ + =

= 2 2 2 2

1 1 1

cos sin cos sin=

= sin2 θ cosec2 θ

Ex.2 (1 + cot2θ) (1 – cos θ) (1 + cos θ) =?

Sol. (1 + cot2 θ) (1 - cos θ) (1 + cos θ)

= cosec2 θ (1 - cos2 θ) = cosec2 θ sin2 θ = 1

Ex.3 cos2 θ cosec θ + sin θ =?

Sol. cos2 θ cosec θ + sin θ

= cos2 θ × 1

sin + sin θ =

2 2cos sin 1

sin sin

+ =

= cosec θ

Ex.4 (1 + tan2 A) cos2 A =?

Sol. (1 + tan2 A) cos2 A

= sec2 A cos2 A [∵ 1 + tan2 A = sec2 A]

= 2

1

cos A× cos2 A = 1

Ex.5 ( cosec2 θ - 1) tan2 θ =?

Sol. (cosec2 θ - 1) tan2 θ

= cot2 θ × tan2 θ [∵ cosec2 θ = 1 + cot2 θ]

= 1 [∵ tan θ × cot θ = 1]

Ex.6 sinA cosA

cosecA secA+ =?

Sol. sinA cosA

cosecA secA+

= sin A × 1

cosecA+ cos A ×

1

secA

= sin A × sin A + cos A × cos A

= sin2 A + cos2 A = 1

Ex.7 1 1

1 cosA 1 cosA+

− + =?

Sol. 1 1

1 cosA 1 cosA+

− +=

( )( )1 cosA 1 cosA

1 cosA 1 cosA

+ + + −

− +

= 2 2

2 2

1 cos A sin A=

− = 2 cosec2 A

Ex.8 sec–tan = 4 then find sec =

Sol. sec - tan = 4 .... (i)

The sec + tan = 1

4 ...... (ii)

Eq.(i) + eq. (ii)

2sec = 4+1

4 ⇒ sec =

17

8

Ex.9 cosec + cot = 3 then find cos =?

Sol. cosec + cot = 3 ...... (i)

Then cosec - cot =1

3 .......... (ii)

(i) + (ii)

2cosec = 3+1

3=

10

3 ⇒ cosec =

5

3

5

4

3

cos = 4

5

b

h=

Examples


48



49

1. (1 + tan2θ) (1 + sin θ) (1 – sin θ)

(a) 0 (b) 1 (c) 2 cosec2θ (d) 2tan2θ sin2θ

2 2

2

1 tan

cosec

+

=?

(a) cot2θ (b) sin2θ (c) cosec2θ (d) tan2θ

3. 2 2

1 1

1 tan 1 cot+

+ +

(a)cosec2θ (b) 0 (c) 2sec2θ (d) 1

4. 2 2

2 2

tan sec

cot cosec

−

− =?

(a) 1 (b) - 4 (c) 4 (d) 7

5. tan2 θ - sin2 θ = ?

(a) tan θ sin θ (b) cot2 θ - cos2 θ

(c) sin2 θ tan2 θ (d) sec2 θ + cos2 θ

6. sin2 A cos2 B - cos2 A sin2 B = ?

(a) sin2A + cos2B (b) sin2A - sin2B

(c) cos2A - sin2B (d) sin2A - cos2B

7. ( )2

2

1 cot tan

sec

+

=?

(a) cos θ (b) cot θ (c) cosec2 θ (d) tan2 θ

8. (cosec2 θ - 1)(sec2 θ - 1) =?

(a) 1 (b) 2 (c) 4 (d) 3

9. cos2 θ + cos2 θ cot2 θ =?

(a) tan2 θ (b) cot2 θ (c) cot2θ (d) tan2θ

10. (cosec θ - sin θ)(sec θ - cos θ)(tan θ + cot θ) = ?

(a) 2 (b) sec2 θ (c) tan2 θ (d) 1

11. sec 1

sec 1

−

+=?

(a) 2

cos

1 cos

+ (b)

2cos

1 sin

−

(c) 2

sin

1 cos

+ (d)

−

21 cos

sin

12. tan θ + cot θ =?

(a) sinθ cos θ (b) 2secθ

(c) secθ cosec θ (d) sec2θ

13. + −

+− +

1 cos 1 cos

1 cos 1 cos

(a) 2sec θ (b) 2cot θ

(c) 2cosec θ (d) cosec θ + cot θ

14. tan2A + cot2A = ?

(a) sec2A + cot2A (b) tan2A + cosec2A - 2

(c) sec2A cosec2A - 1 (d) sec2A cosec2A - 2

15. 1 1

sec 1 sec 1−

− +=?

(a) cot2θ (b) 2sec2 θ (c) 2

2

cot (d)

2

2

tan

16. cot2A cosec2B - cot2B cosec2A = ?

(a) cot2A - cot2B (b) cosec2A - cot2B

(c) tan2A + sin2B (d) cot2A + cot2B

17. tan sin

tan sin

+

− =?

(a) sec 1

sec 1

+

− (b)

cosec 1

sec

+

(c) sec

cosec 1

− (d) 1

18. 1 sin

1 sin

−

+ =?

(a) (sec θ + tan θ)2 (b) (sec θ - tan θ)2

(c) (cosec θ - cot θ)2 (d) (cosec θ - cot θ)2

19. 2 2

1 1

cos sin+

=?

(a) sin2 θ cos2 θ (b) 2cosec2 θ

(c) tan2 θ sin2 θ (d) sec2 θ cosec2 θ

Exercise - 1


50

20. sec4 θ - sec2 θ =?

(a) tan4 θ - tan2 θ (b) tan2 θ - tan4 θ

(c) tan4 θ + tan2 θ (d) None of these

21. cos4 A + sin2 A cos2 A =?

(a) sec2A (b) cos2A (c) sin2A (d) cos2A

22. sinA cosA sinA cosA

sinA cosA sinA cosA

+ −+

− +

(a) 2

2

2sin A 1− (b)

2

2

2cos A 1−

(c) 0 (d) 1

23. 2 2sec cosec+ =

(a) tan θ + cot θ (b) sin θ + cos θ

(c) cos θ + cot θ (d) tan θ + cos θ

24. sin6 A + cos6 A

(a) 1 + 3sin2A cos2A (b) 1 - 3sin2A + 4cos2A

(c) 1 + 3sin2A - 4cos2A (d) 1 - 3sin2A cos2A

25. cot A tanB

cot B tan A

+

+=?

(a) cot A + tan B (b) cot A tan B

(c) cos B – tan A (d) cot B + tan B



51

1. If tan + cot = 2 then value of tan5+cot5 is

(a) 2 (b) 1 (c) 10 (d) 10

2. If sin + cosec = 2 then the value of sinn + cosecn is

(a) 2n (b) 2n (c) 2 (d) 2n+1

3. If cos + sec = 2, then the value of cos9 + sec9 is

(a) 2 (b) 29 (c) 1 (d) 2

4. If tan + cot = 2 then the value of tan-10 + cot-15 =?

(a) -2 (b) 2 (c) 210 (d) 2-5

5. If sin2+sin2 = 2 then cos 2

+

=?

(a) 1 (b) -1 (c) 0 (d) 0.5

6. if sec + tan = 2 then sec = ?

(a) 5

2 (b)

5

4 (c)

7

4 (d)

7

2

7. If cot A + cosec A = 3 and A is acute angle then value of

cot A is

(a) 4

5 (b) 1 (c)

1

2 (d)

3

4

8. If sec + tan = 2+ 5 then value of sin + cos

(a) 5 (b) 7

5 (c)

1

5 (d)

3

5

9. If cosec = x+1

4x (0ᵒ<<90ᵒ) then value of cosec + cot

is

(a) 2

x (b) 2x (c) x (d)

1

4x

10. If 3sin+5cos = 5 then 5 sin - 3cos =?

(a) ±3 (b) ±5 (c) 1 (d) ±2

11. sin - cos = 7

13and 0ᵒ < < 90ᵒ then sin + cos

equals

(a) 17

13 (b)

13

17 (c)

1

13 (d)

1

17

12. If sec2 + tan2 = 7 then the value of is

(a) 60ᵒ (b) 30ᵒ (c) 0ᵒ (d) 90ᵒ

13. If be an acute angle and 7sin2 + 3cos2 = then the

value of tan is

(a) 3 (b) 1

3 (c) 1 (d) 0

14. Find value of (cos+sin)2+ (cos-sin)2

(a) 2 (b) 3 (c) 4 (d) 5

15. If sec2 + tan2 = 7

12then sec4 - tan4 =?

(a) 49

144 (b)

12

7 (c)

7

12 (d) 1

16. sec4 - sec2 = ?

(a) tan2 + cot2 (b) tan4 + cot2

(c) tan4 + tan2 (d) sec2 + cosec2

17. If x = a (sin + cos), y = b (sin - cos) then the value of 2 2

2 2

x y

a b+ is

(a) 0 (b) 1 (c) 2 (d) -2

18. If x = sinA cosB, y = sinAsinB, & z = cosA then

(a) x2 + y2 + z2 = 2 (b) x2 - y2 + z2 = 2

(c) x2 + y2 - z2 = 2 (d) -x2 + y2 + z2 = 2

19. If cos + cos2 = 1 then sin8 + 2sin6+sin4 =?

(a) 0 (b) -1 (c) 1 (d) 2

20. if cot + tan = x and sec - cos = y, then

( ) ( )2 3

2 23 2x y xy− = ?

(a) 4 (b) 3 (c) 2 (d) 1

21. if tan - tan2 = 1 then the value of sec2 - sec4 is

(a) 1 (b) -1 (c) 2 (d) 0

Exercise - 2


52

22. the value of 1 1

cos cot sinec −

− is

(a) 1 (b) cot (c) cosec (d) tan

23. cosec - cot = 7

2, then cot = ?

(a) 47

28 (b)

83

28 (c)

49

28− (d)

45

28−

24. If cos2 - sin2 = 1

3 then find the value of (cos4 -

sin4+1)?

(a) 1 (b) 1

3 (c)

4

3 (d)

5

3

25. Value of cot180 (cot720 cos2220 +0 2 0

1

tan72 sec 68) is

(a) 1 (b) 2 (c) 3 (d) 1

3



53

1. tan sec 1

tan sec 1

+ −

− +=?

(a) 1 sin

cos

+

(b)

1 cos

sin

+

(c)

1 tan

cos

+

(d) 1

2. cot A cosecA 1

cot A cosecA 1

+ −

− +

(a) 1 sinA

cosA

− (b)

1 cosA

sinA

+ (c)

1 cotA

cosA

− (d) 1

3. sec A (1 – sin A) (sec A + tan A)

(a) - 1 (b) 1 (c) 2 (d) - 3

4. 2cosA sin A

1 tan A sin A cosA+

− −

(a) sinA - cosA (b) sin2A - cos2A

(c) sinA + cosA (d)3sinA + 2cosA

5. 2 3cos sin

1 tan sin cos

+

− −

(a) 1+ sin θ + cos θ (b) 1-sin θ cos θ

(c) 1+sin θ cos θ (d) 1 + sin θ - cos θ

6. 3 3sin cos

sin cos

+

+ + sin θ cos θ

(a) 0 (b) 1 (c) 2 (d) 3

7. ( sin A + sec A)2 + (cos A + cosec A)2 = ?

(a) (1 – sec A cosec A)2 (b) (1 + sec A cosec A)2

(c) (1 + sin A + cos A)2 (d) (1 + sec A - cosec A)2

8. (1 + tan A tan B)2 + (tan A - tan B)2=?

(a) sec2A sec2B (b) sec2A + sec2B

(c) sec2 A - sec2 B (d) sec4 A sec2 B

9. sin θ (1 + tan θ) + cos θ (1 + cot θ) =?

(a) cosec θ - sec θ (b) sec θ - cosec θ

(c) cosec θ + sec θ (d) 1

10. tan6 θ + 3 tan2 θ sec2θ + 1 =?

(a) sec6 θ (b) sin6 θ (c) sec8 θ (d) cosec6 θ

11. 2(sin6 θ + cos6 θ) – 3(sin4 θ + cos4 θ) + 1

(a) 0 (b) 1 (c) - 2 (d) 3

12. tan cot

sin cos

−

=?

(a) tan2 θ + cot2 θ (b) tan2 θ - cot2 θ

(c) cos2 θ - cosec2 θ (d) sin2 θ + cot2 θ

13. sec4A (1-sin4A) - 2 tan2A=?

(a) 2 (b) 3 (c) 5 (d) 1

14. secA tanA

secA tanA

−

+=?

(a) 1 - 2 sec A tanA + 2 tan2 A

(b) 1 + 2 sec A tanA + 2 tan2 A

(c) 1 + 2 sec A tanA - 2 tan2 A

(d) 1 - 2 sec A tanA - 2 tan2 A

15. 2 sec2 θ - sec4 θ - 2 cosec2 θ + cosec4 θ

(a) cot4 θ - tan4 θ

(b) cot4 θ + tan4 θ

(c) cosec4 θ - tan4 θ

(d) cos4 θ + cosec4 θ

16. − +

++ −

1 sin 1 sin

1 sin 1 sin

(a) 2 tan θ (b) 2 cosec θ

(c) sec θ - tan θ (d) 2sec θ

17. 2 cosec2230 cot2670 - sin2230 - sin2670 - cot2670 =?

(a) 1 (b) sec2 230 (c) tan2 230 (d) 0

18. sin cos

1 cot 1 tan

+

− − =?

(a) 0 (b) 1

(c) sin θ + cos θ (d) sin θ - cos θ

19. Value of cos cos

1 sin 1 sin

+

− + is

(a) 2 sec θ (b) sec2 θ (c) 2 cosec θ (d) 2 sin θ

20. sinA 1 cosA

1 cosA sinA

++

+=

(a) 2sinA (b) 2 cosecA (c) tan2

A (d) cot

2

A

Exercise - 3


54

21. (1 + cot θ - cosec θ)(1 + tan θ + sec θ) =?

(a) 3 (b) 1 (c) 0 (d) 2

22. tan tan

sec 1 sec 1

−

− +=?

(a) 2sin (b) 2sec

(c) 2 cosec (d) cosec + sec

23. tan2 𝜙 + cot2 𝜙 + 2 =?

(a) 2 (b) sec2 𝜙 sin2 𝜙

(c) cos2 𝜙 cosec2 𝜙 (d) sec2 𝜙 cosec2 𝜙



55

Answer key 1 b 2 d 3 d 4 a 5 c

6 b 7 b 8 a 9 b 10 d

11 d 12 c 13 c 14 d 15 d

16 a 17 a 18 b 19 d 20 c

21 d 22 a 23 a 24 d 25 b

1. (b)

(1 + tan2 θ) (1 + sin θ) (1 - sin θ)

= (1 + tan2 θ) (1 - sin2 θ)

= sec2 θ cos2 θ = 2

1

cos cos2 θ = 1

2. (d)

2

2

1 tan

cosec

+

=

2 2 2

2 2

2 2

sin cos sin1

cos cos1 1

sin sin

+ +

=

= 22

2

2

1sincos

1 cos

sin

=

= tan2 θ

3. (d)

2 2

1 1

1 tan 1 cos+

+ + =

2 2

1 1

sec cosec+

= cos2 θ + sin2 θ = 1

4. (a)

2 2

2 2

tan sec

cot cosec

−

− =

( )( )

2 2

2 2

sec tan

cosec cot

− −

− − =

1

1

−

−= 1

5. (c)

tan2 θ - sin2 θ = 2

2

sin

cos

- sin2 θ

= sin2 θ 2

11

cos

−

= sin2 θ

( )2

2

1 cos

cos

−

= sin2 θ 2

2

sin

cos

= sin2 θ tan2 θ

6. (b)

sin2A cos2B - cos2Asin2B

= sin2A (1-sin2B) - (1-sin2A) sin2B

= sin2A - sin2A sin2B - sin2B + sin2A sin2B

= sin2A - sin2B

7. (b)

(cosec2 θ - cot2 θ) = 1

⇒ 1+cot2 θ = cosec2 θ) =

2

2

cosec tan

sec

=

2

2

1 sin

cossin1

cos

=

2

2

sin cos cos

sinsin cos

=

= cot θ

8. (a)

(cosec2 θ - 1)(sec2 θ - 1)

= cot2 θ tan2 θ =

2 2

2 2

cos sin

sin cos = 1

9. (b)

cos2 θ + cos2 θ cot2 θ = cos2 θ (1 + cot2 θ)

= cos2 θ cosec2 θ = cos2 θ2

1

sin =

2

2

cos

sin

= cot2 θ

10. (d)

(cosec θ - sin θ)(sec θ - cos θ)(tan θ + cot θ)

= 1 1 sin cos

sin cossin cos cos sin

− − +

= 2 2 2 21 sin 1 cos sin cos

sin cos sin cos

− − +

=

2 2

2 2

cos sin 1

sin cos= 1

11. (d)

sec 1

sec 1

−

+ =

11

cos1

1cos

−

+

= −

+

1 cos

1 cos

= ( )( )

( )( )

− −

+ −

1 cos 1 cos

1 cos 1 cos =

( )2 2

2

1 cos 1 cos

sinsin

− − =

12. (c)

Solutions

Exercise - 1


56

tan θ + cot θ = sin cos

cos sin

+

= 2 2sin cos 1

cos sin cos sin

+ =

= 1 1

cos sin

= secθ cosec θ

13. (c)

+ −

+− +

1 cos 1 cos

1 cos 1 cos

= ( )( )( )( )

1 cos 1 cos

1 cos 1 cos

+ +

− + + ( )( )

( )( )

1 cos 1 cos

1 cos 1 cos

− −

+ −

= ( ) ( )+ −

+− −

2 2

2 2

1 cos 1 cos

1 cos 1 cos

= 1 cos 1 cos

sin sin

+ − +

=

2

sin= 2 cosec θ

14. (d)

tan2A + cot2A = 2 2

2 2

sin A cos A

cos A sin A+

= 4 2

2 2

sin A cos A

sin Acos A

+

= ( )

22 2 2 2

2 2

sin A cos A 2sin Acos A

sin Acos A

+ −

= 2 2

2 2

1 2sin Acos A

sin Acos A

− = cosec2A sec2A – 2

15. (d)

1 1

sec 1 sec 1−

− + =

( )( )( )

− − −

+ −

sec 1 sec 1

sec 1 sec 1

= + − +

−2

sec 1 sec 1

sec 1 =

2

2

tan

[∵ sec2 θ - 1 = tan2 θ]

16. (a)

cot2A cosec2B - cot2B cosec2A

= cot2A (1+ cot2B) - cot2B (1 + cot2A)

= cot2A + cot2Acot2B - cot2B - cot2Acot2B

= cot2A - cot2B

17. (a)

tan sin

tan sin

+

− =

sinsin

cossin

sincos

+

−

=

11

sec 1cos1 sec 1

1cos

+ + = −

−

18. (b)

1 sin

1 sin

−

+

( )( )

1 sin

1 sin

−

−

= ( )

22

2 2

1 sin 1 sin 2sin

1 sin cos

− + − =

−

= sec2 θ + tan2 θ - 2 sec θ tan θ = (sec θ - tan θ)2

19. (d)

2 2

1 1

cos sin+

=

2 2

2 2 2 2

sin cos 1

cos sin cos sin

+ =

= sec2 θ cosec2 θ

20. (c)

sec4 θ - sec2 θ = sec2 θ (sec2 θ - 1) = (1 + tan2 θ) tan2 θ

= tan2 θ + tan4 θ = tan4 θ + tan2 θ

21. (d)

cos4 A + sin2 A cos2 A

= cos2 A(cos2 A + sin2 A) = cos2 A × 1 = cos2 A

22. (a)

sinA cosA sinA cosA

sinA cosA sinA cosA

+ −+

− +

( ) ( )( )( )

2 2sinA cosA sinA cosA

sinA cosA sinA cosA

+ + −

− +

( )

= =− − −

2 2 2 2

2 2

sin A cos A sin A 1 sin A=

2

2

2sin A 1−

23. (a)

2 2sec cosec+

= +

+ =

2 2

2 2 2 2

1 1 sin cos

cos sin sin cos

= 1

sin cos =

2 2sin cos

sin cos

+

= tan θ + cot θ

24. (d)

sin6A + cos6A = (sin2 A)3 + ( cos2 A)3

= (sin2A + cos2A)3 – 3sin2A cos2A (sin2A + cos2A)

= 1-3sin2Acos2A

25. (b)

cot A tanB

cot B tan A

+

+ =

1tanB

tan A1

tan AtanB

+

+

= ( )

( )

1 tanBtan A tanB

tan A 1 tan A tanB

+

+ =

tanB

tan A= cot A tan B



57

Answer key 1 a 2 c 3 a 4 b 5 c

6 b 7 a 8 d 9 b 10 a

11 a 12 a 13 b 14 a 15 c

16 c 17 c 18 a 19 c 20 d

21 a 22 b 23 d 24 c 25 a

1. (a)

tan + cot = 2

⇒ tan +1

tan = 2 ⇒ tan2 + 1 = 2 tan

⇒ (tan - 1)2 = 0 ⇒ tan = 1

∴ cot = 1

tan5+cot5 = 15+15 = 2

2. (c)

sinn + cosecn ⇒ sin + 1

sin = 2

⇒ sin2+1 = 2sin ⇒ (sin−)2 = 0

⇒ sin = 1 ⇒ cosec = 1

[ Now sinn + cosecn = 1n+1n = 2 ]

3. (a)

cos + sec = 2

cos + 1

cos = 2 cos2 + = cos

⇒ (cos − ) = 1 cos =

∴ cos9 + sec9 = 1 + 1 = 2

4. (b)

tan + cot = 2 ⇒ tan + 1

tan = 2

tan2 + = tan ⇒ (tan - 1)2 = 0 tan =

Now tan-10 + cot-15 = 10 15

1 1

tan cos + =

1 1

1 1+ = 2

5. (c)

sin2 + sin2 = 2

Possible if and only if sin = sin = 1

∴ = = 90ᵒ

cos2

+

= cos90 90

2

+

= cos90ᵒ = 0

6. (b)

sec+tan = 2 ...... (i)

∴ sec - tan = 1

2 ......... (ii)

(i)+(ii) 2sec = 5

2 ⇒ sec =

5

4

7. (a)

cosec A + cot A = 3 ........ (i)

⇒ cosec A – cot A = 1

3 ........... (ii)

(i) + (ii)

2cosec A = 10

3 ⇒ cosec A =

5

3

5

3

4

A

cos A = 4

3

8. (d)

sec + tan = 5 + 2 ........... (i)

⇒ sec - tan = 1

5 2+ = 5 - 2 ..... (ii)

(i) + (ii)

2sec = 2 5 ⇒ sec = 5

Now

2

1

sin = 2

5, cos =

1

5

sin + cos = 2

5+

1

5 =

3

5

9. (b)

cosec = x+1

4x

Exercise - 2


58

let cosec + cot = k ................. (i)

cosec - cot = 1

k .................. (ii)

(i) +(ii)

2 cosec = k+1

k ⇒ cosec =

1 1

2k

k

+

by comparison

1 1

2k

k

+

= x+

1

4x =

1 12

2 2x

x

+

⇒ k = 2x, 1

2x

cosec + cot = x

10. (a)

3sin +5cos = 5 ........... (i)

And let 5sin - 3cos = k .......... (ii)

(i)2 + (ii)2

32+52 = 52 + k2 ⇒ k = ±3

11. (a)

Let sin + cos = k ….(i)

sin - cos = 7

13 .... (ii)

(i)2+(ii)2

1+1 = k2+49

169 ⇒ k2 = 2-

49

169 =

289

169 ⇒ k = ±

17

13

12. (a)

sec2 + tan2 = 7 ⇒ (1 + tan2 ) + tan2 = 7

⇒ 2tan2 = 6 ⇒ tan2 = 3 ⇒ tan = 3

∴ = 60ᵒ

13. (b)

7 sin2+3cos2 = 4 ⇒ 7 sin2 + 3 (1-sin2) = 4

⇒ 7 sin2 + 3 – 3 sin2 = 4 ⇒ 4 sin2 = 1

⇒ sin2 =1

4 ⇒ sin =

1

2

= 30ᵒ

∴ tan 30ᵒ = 1

3

14. (a)

(cos+sin)2 + (cos-sin)2

= cos2 + sin2 + 2sincos + cos2+sin2 - 2sincos

= 2(sin2 + cos2) = 2

15. (c)

sec4 - tan4 = (sec2 - tan2) (sec2+tan2)

= 1×7

12 =

7

12

16. (c)

sec4 - sec2 = sec2(sec2−) = sec2 (tan2)

= (1+tan2)(tan2) = tan2 + tan4

17. (c)

x = a (sin + cos)

⇒ x

a = sin +cos ….. (i)

⇒ y = b (sin-cos)

⇒ y

b = sin - cos ..... (ii)

(i)2+(ii)2

2 2

2 2

x y

a b+ = (sin + cos)2 + (sin - cos)2

= sin2 + cos2 + 2sin cos + sin2 + cos2 -

2sin cos =

18. (a)

x2 + y2 + z2 = (sinAcosB)2+(sinAsinB)2+(cosA)2

= 2sin2Acos2B + 2sin2Asin2B + 2cos2A

= 2sin2A (cos2B+sin2B)+2cos2A

= sin2A + 2cos2A = 2 (sin2A+cos2A) = 2

19. (c)

cos = 1 – cos2

cos = sin2 = sin8 + 2sin6 + sin4

= cos4 + 2cos3 + cos2

= (cos2 + cos)2 = (1)2 = 1

20. (d)

cot + tan = x

cos sin

sin cos

+ = x ⇒

2 2cos sin

sin cos

+ = x

⇒ x = 1

sin cos and sec - cos = y

1

cos - cos = y ⇒

21 cos

cos

− = y ⇒ y =

2sin

cos

Now

( ) ( )2 3

2 23 2x y xy− =

2 32 43 2

2 2 2

1 sin 1 sin

cos sin cossin cos cos

−

= 1

= ( ) ( )2 2

3 33 3sec tan − = sec2 - tan2 = 1

21. (a)

tan - tan2 = 1

⇒ tan = 1+tan2 = sec2



59

∴ sec2 - sec4 = tan - tan2 = 1

22. (b)

1 1

cos cot sinec −

−

= cosec + cot - cosec = cot

23. (d)

cosec - cot = 7

2 ............. (i)

then cosec+cot = 2

7 ........... (ii)

(ii)-(i)

2cot = 2

7-

7

2 =

4 49

14

− = -

45

14 ⇒ cot = -

45

28

24. (c)

cos2 - sin2 = 1

3

(cos4 - sin4)+1

= (cos2-sin2)(cos2+sin2)+1 = 1 4

13 3

+ =

25. (a)

cot180 (cot720 cos2220 +0 2 0

1

tan72 sec 68)

= cot180 (tan 180 cos2220 +0 2 0

1

cot18 cosec 22)

= cot180 (tan180 cos2220 + tan 180 sin2220)

= cot180 tan180 (cos2220 + sin2220) = 1 × 1 = 1

Answer key

1 a 2 b 3 b 4 c 5 c

6 b 7 b 8 a 9 c 10 a

11 a 12 b 13 d 14 a 15 a

16 d 17 b 18 c 19 a 20 b

21 d 22 c 23 d

1. (a)

tan sec 1

tan sec 1

+ −

− + =

( ) + −

− −

sec tan 1

1 sec tan

(Let sec θ + tan θ = x

then sec θ – tan θ = 1

x) =

− −=

−−

x 1 x 1

1 x 11

x x

= x

= sec θ + tan θ = 1 sin

cos cos

+

=

1 sin

cos

+

2. (b)

cot A cosecA 1

cot A cosecA 1

+ −

− + =

( )( )

cosecA cot A 1

1 cosecA cot A

+ −

− −

(Let cosec A + cot A = x

then cosec A - cot A = 1

x) =

x 1 x 1

1 x 11

x x

− −=

−−

= x

= cosec A + cot A = +1 cosA

sinA sinA =

1 cosA

sinA

+

3. (b)

sec A (1 – sin A) (sec A + tan A)

= sec A (1 - sin A) 1 sinA

cosA cosA

+

= sec A (1-sinA) ( )1 sin A

cos A

+=

2 2

2 2

1 sin A cos A

cos A cos A

−= = 1

4. (c)

2cosA sin A

1 tan A sin A cosA+

− − = +

−−

2cosA sin A

sinA sinA cosA1

cosA

= 2 2cos A sin A

cosA sin A sin A cosA+

− −

= 2 2sin A cos A

sin A cosA sin A cosA−

− −

Exercise - 3


60

= 2 2sin A cos A

sin A cos A

−

−= sinA + cosA

5. (c)

2 3cos sin

1 tan sin cos

+

− − =

+

− −

2 3cos sin

sin sin cos1

cos

= 3 3cos sin

cos sin sin cos

+

− − =

−

− −

3 3sin cos

sin cos sin cos

= 3 3sin cos

sin cos

−

− =

( )( )( )

2 2sin cos sin cos sin cos

sin cos

− + +

−

= (1 + sin θ cos θ)

6. (b)

( )( )( )

+ + −

+

2 2sin cos sin cos sin cos

sin cos+sinθcos θ

= 1 - sin θ cos θ + sin θ cos θ = 1

7. (b)

sin2A + sec2A + 2sinAsecA + cos2A + cosec2A +

2cosAcosecA

= 1 + sec2A + cosec2A + 2sinA cosA

cosA sinA

+

= 1 + sec2A + cosec2A + 2 2 2sin A cos A

sin AcosA

+

= 1 + sec2A + cosec2A + 2sec A cosec A

= 1 + 2 2

1 1

cos A sin A+ + 2sec A cosec A

= 1 + 2 2

2 2

sin A cos A

sin Acos A

++ 2sec A cosec A

= 1 + sec2 A cosec2 A + 2sec A cosec A

= (1 + sec A cosec A)2

8. (a)

(1 + tanA tanB)2 + (tanA - tan B)2

= 1 + tan2A tan2B + 2tanA tanB + tan2A + tan2B - 2tan A

tan B

= 1 + tan2 A tan2 B + tan2A + tan2 B

= (1+tan2 A) (1+tan2B) = sec2 A sec2 B

9. (c)

sin θ (1 + tan θ) + cos θ (1 + cot θ)

= sin θsin

1cos

+

+ cos θ

cos1

sin

+

= sin θ cos sin

cos

+

+ cos θsin cos

sin

+

= (sin θ + cos θ) sin cos

cos sin

+

=(sin θ + cos θ)2 2sin cos

sin cos

+

= (sin θ + cos θ)1

sin cos

= sin cos

sin cos sin cos

+

=

1 1

cos sin+

= cosec θ + sec θ

10. (a)

= tan6 θ + 3 tan2 θ sec2θ + 1

= tan6 θ + 3 tan2 θ (1 + tan2 θ) + 1

= (tan2θ)3 + 3 tan2 θ (1 + tan2 θ) + 13

= (tan2 θ + 1)3 = (sec2 θ)3 = sec6 θ

11. (a)

2(sin6 θ + cos6 θ) - 3(sin4 θ + cos4 θ) + 1

= 2 [(sin2 θ + cos2 θ)3 - 3sin2 θ cos2 θ (sin2 θ + cos2 θ)] - 3

[(sin2 θ + cos2 θ)2 - 2sin2 θ cos2 θ] + 1

= 2 [1 - 3 sin2 θ cosec2 θ] - 3 (1 - 2sin2 θ cos2 θ) + 1

= 2 - 6 sin2 θ cos2 θ - 3 + 6 sin2 θ cos2 θ + 1 = 0

12. (b)

tan cot

sin cos

−

=

sin cos

cos sinsin cos

−

=

2 2

2 2

sin cos

sin cos

−

= 2 2

1 1

cos sin−

= sec2 θ - cosec2 θ

= 1 + tan2 θ - (1 + cot2 θ) = tan2 θ - cot2 θ

13. (d)

sec4A (1-sin4A) - 2 tan2A

= sec4A cos2A (1+sin2A) - 2tan2A

= sec4A 2

1

sec A(1+sin2A) - 2tan2A

= sec2A (1+sin2A) - 2tan2A

= sec2A + tan2A - 2 tan2A = sec2A - tan2A = 1

14. (a)

secA tanA

secA tanA

−

+ =

( )( )( )( )

secA tan A secA tan A

secA tan A secA tanA

− −

+ −

= ( )

2

2 2

sec A tan A

sec A tan A

−

−=

2 2sec A tan A 2secAtanA

1

+ −

= 1 + tan2 A + tan2 A - 2 sec A tanA

= 1 + 2 tan2 A - 2sec A tan A

= 1 - 2 sec A tan + 2 tan2 A



61

5. (a)

2 sec2 θ – sec4 θ - 2 cosec2 θ + cosec4 θ

= 2 (1+tan2θ) - (1+tan2θ)2 - 2(1+cot2θ) + (1+cot2 θ)2

= 2 + 2 tan2 θ – (1 + tan4 θ + 2tan2 θ) - 2 - 2 cot2 θ + (1 +

cot4 θ + 2 cot2 θ) = cot4 θ - tan4 θ

16. (d)

− +

++ −

1 sin 1 sin

1 sin 1 sin

= ( )( )( )( )

− −

+ −

1 sin 1 sin

1 sin 1 sin+ ( )( )

( )( )

+ +

− +

1 sin 1 sin

1 sin 1 sin

= ( ) ( )− +

+− −

2 2

2 2

1 sin 1 sin

1 sin 1 sin

= ( ) ( )− +

+

2 2

2 2

1 sin 1 sin

cos cos

= 1 sin 1 sin

cos cos

− + +

= (sec θ - tan θ) + (sec θ + tan θ)

= 2 sec θ

17. (b)

2 cosec2230 cot2670 - sin2230 - sin2670 - cot2670

= (2 cosec2230 - 1) cot2670 - sin2230 - sin2670

= cot2670 [cosec2230 + cosec2230 - 1] - (sin2230 +

cos2230)

= tan 2230 [cosec2230 + cot2230] - 1

= tan2230 × cosec2230 + tan2230cot2230 -1

= 2 0

2 0 2 0

sin 23 1

cos 23 sin 23+ 1 – 1 =

2 0

1

cos 23= sec2230

18. (c)

sin cos

cos sin1 1

sin cos

+

− −

= 2 2sin cos

sin cos cos sin

+

− −

= 2 2sin cos

sin cos sin cos

−

− −

= 2 2sin cos

sin cos

−

− = sin θ + cos θ

19. (a)

cos cos

1 sin 1 sin

+

− +

=( )

( )( )( )

( )( )

cos 1 sin cos 1 sin

1 sin 1 sin 1 sin 1 sin

+ − +

− + + −

= ( ) ( )

2 2

cos 1 sin cos 1 sin

1 sin 1 sin

+ − +

− −

= 1 sin 1 sin

cos cos

+ − +

=

1 sin 1 sin

cos

+ + −

=2

cos= 2 sec θ

20. (b)

sinA 1 cosA

1 cosA sinA

++

+ =

( )( )

22sin A 1 cosA

sinA 1 cosA

+ +

+

= ( )

2 2sin A 1 cos A 2cosA

sinA 1 cosA

+ + +

+

= ( )

2 2sin A cos A 1 2cosA

sinA 1 cosA

+ + +

+

= ( ) ( )

1 1 2cosA 2 2cosA

sinA 1 cosA sinA 1 cosA

+ + +=

+ +

= ( )

( )

2 1 cosA

sin A 1 cosA

+

+ =

2

sinA= 2cosec A

21. (d)

(1 + cot θ - cosec θ)(1 + tan θ + sec θ)

= cos 1 sin 1

1 1sin sin cos cos

+ − + +

= sin cos 1 cos sin 1

sin cos

+ − + +

= ( )

2 2sin cos 1

sin cos

+ −

=

2 2sin cos 2sin cos 1

sin cos

+ + −

= 1 2sin cos 1

sin cos

+ −

=

2sin cos

sin cos

= 2

22. (c)

tan tan

sec 1 sec 1

−

− + = tan θ

1 1

sec 1 sec 1

+

− +

= tan θ( )( )sec 1 sec 1

cos 1 sec 1

+ + −

− +

=2sec 2sec 2sec cos

sintan sin

cos

= =

= 2

sin [∵ sec θ cos θ = 1]

= 2 cosec θ

23. (d)

tan2 𝜙 + cot2 𝜙 + 2

= tan2 𝜙 + cot2 𝜙 + 2 tan 𝜙 × cot 𝜙 [∵ tan 𝜙 cot 𝜙 = 1]

= (tan 𝜙 + cot 𝜙)2 =

2sin cos

cos sin

+

=

2 22sin cos 1

cos sin cos sin

+ =

=

21 1

cos sin

= (sec 𝜙 cosec𝜙)2

= sec2 𝜙 cosec2 𝜙

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Trigonometry

Sum and difference of angles

of T-Ratio

Chapter - 06


62

* sin (A + B) = sin A cos B + cos A sin B

* sin (A – B) = sin A cos B – cos A sin B

* cos (A + B) = cos A cos B – sin A sin B

* cos (A – B) = cos A cos B + sin A sin B

* tan (A + B) = ( )( )

sin A B

cos A B

+

+

= sinAcosB cosAsinB

cosAcosB sinAsinB

+

−

= tanA tanB

1 tanAtanB

+

−

* tan (A – B) = tanA tanB

1 tanAtanB

−

+

* cot (A + B) = cot Acot B 1

cot B cot A

−

+

* cot (A – B) = cot Acot B 1

cot B cot A

+

−

* sin (A + B) sin (A – B)

= (sin A cos B + cos A sin B) (sin A cos B – cos A sin B)

= sin2 A cos2 B – cos2 A sin2 B

= sin2 A (1 – sin2 B) – (1 – sin2 A) sin2 B

= sin2 A – sin2 A sin2 B – sin2 B + sin2 A sin2 B

= sin2 A – sin2 B

= 1 – cos A –( A– cos2 B)

= cos2 B–cos2 A

*sin (A + B) sin (A – B) = sin2 A – sin2 B = cos2 B – cos2 A

* cos (A + B) cos (A – B) = cos2 B – sin2 A = cos2 A - sin2 B

* tan (A + B + C)

= tan [(A + B) + C]

=

++

−

+ −

−

tan A tanBtanC

1 tan A tanBtan A tanB

1 tanC1 tan A tanB

= + + −

− − −

tanA tanB tanC tanAtanBtanC

1 tanAtanB tanBtanC tanCtanA

* Values at 15°, 75°

sin15°

= sin(45° – 30°)

= sin45° cos30° – cos45°sin30°

= 1 3 1

2 22 2

− = 3 1

2 2

−

cos15°

= cos(45° – 30°)

= cos45° cos30° – sin45°sin30°

= + 1 3 1 1

2 22 2 = 3 1

2 2

+

sin75°

= sin(90° - 15°) = cos15° = 3 1

2 2

+

cos75°

= cos(90° - 15°) = sin15° = −3 1

2 2

sin15° = cos75° = −3 1

2 2

sin75° = cos15° = 3 1

2 2

+

tan15°

= tan (45° – 30°) = tan45 tan30

1 tan45 tan30

−

+

= ( )( )

( )( )

− − −=

+ −+

11 3 1 3 13

1 3 1 3 113

= 3 1 2 3

2 32

+ −= −

tan75°

= tan(45° + 30°) = tan45 tan30

1 tan45 tan30

+

−

=

13

13

+

−

= ( )( )

3 13 1

3 1 3 1

++

− − =

3 1 2 32 3

2

+ += +

cot15° = cot(90° – 75°) = tan75° = 2 3+

cot75° = cot(90° – 15°) = tan15° = 2 3−

tan15° = cot75° = 2 3−

tan75° = cot15° = 2 3+

Sum and Difference of angles of

T - Ratio 6



63

1 Value of ( )sin A B

cosAcosB

− =?

Sol. ( )−sin A B

cosAcosB =

−sinAcosB cosAsinB

cosAcosB

= sinAcosB cosAsinB

cosAcosB cosAcosB− = tan A – tan B

2 If sin A = 3

5 and cos B =

12

13 where (0° A, B 90°) then

find value of

(i) sin (A + B) (ii) cos (A + B)

Sol. sin A = 3

5

cos A =

23 4

15 5

− =

cos B = 12

13

sin B =

212 5

113 13

− =

sin (A + B) = sin A cos B + cos A sin B

= 3 12 4 5

5 13 5 13 + =

36 20 56

65 65 65+ =

cos (A + B) = cos A cos B – sin A sin B

= 4 12 3 5

5 13 5 13 − =

33

65

3 If (A + B) = 45° then

(i) tan A + tan B + tan A tan B = ?

(ii) (1 + tan A) (1 + tan B) =?

Sol. (i) A + B = 45°

tan (A + B) = tan45° tanA tanB

1 tanAtanB

+

− = 1

tan A + tan B = 1 – tan A tan B

tan A + tan B + tan A tan B = 1

(ii) tan A + tan B + tan A tan B = 1

1 + tan A + tan B + tan A tan B = 1 + 1

(1 + tan A) + tan B(1 + tan A) = 2

(1 + tan A)(1 + tan B) = 2

4 If A + B = 45° then (1 – cot A)(1 – cot B) = ?

Sol. A + B = 45°

cot (A + B) = cot 45° cot Acot B 1

cot A cot B

−

+ = 1

cot A cot B – 1 = cot A + cot B

1 + cot A cot B – cot A – cot B = 1 + 1

1 – cot A – cot B + cot A cot B = 2

1 – cot A – cot B(1 – cot A) = 2

(1 – cot A)(1 – cot B) = 2

5 If tan x tan y = 1

3 then 2 cos (x + y) =?

Sol. tan x tan y = 1

3

sinx siny 1

cosx cosy 3=

Using componendo and dividendo

cosxcosy sinxsiny 3 1

cosxcosy sinxsiny 3 1

− −=

+ +

( )( )

cos x y 1

2cos x y

+=

− 2 cos (x + y) = cos (x – y)

6 If x = sec2 + cosec2 and y = cot2 + tan2 then 1 + x –

y = ?

Sol. 1 + x – y

= 1 + sec2 + cosec2 – cot2 + tan2 = 1 + 1 + 1 = 3

7 sin 15° + cos 15° = ?

Sol sin 15° + cos 15°

= 3 1 3 1

2 2 2 2

− ++ =

3 1 3 1

2 2

− + + =

3

2

Examples


64

1. tan (45°+A) =?

(a) 1 tan

1 tan

A

A

+

− (b) 1

(c) 1 tan

1 tan

A

A

−

+ (d)

2tan

1 tan

A

A+

2. cot(45° - A) =?

(a) 1 cot

1 cot

A

A

+

− (b)

cot 1

cot 1

A

A

−

+

(c) cot 1

cot 1

A

A

+

− (d) 1

3. ( )( )

tan tan

1 tan tan

A B B

A B B

+ −

+ −=?

(a) tan (A+2B) (b) tan A

(c) tan B (d) 1

4. tan30 tan15

tan30 tan15 1

+

−

(a) 1 (b) -1 (c) 0 (d) tan 150

5. Find tan (A + B) if tan A = 1

2 and tan B =

1

3

(a) 5

6 (b) -1 (c)

6

5 (d) 1

6. sin22 cos22

sin22 cos22

+

− =?

(a) tan 67° (b) – tan 67°

(c) tan 23° (d) – tan 23°

7. tan69 tan66

11 tan69 tan66

+ +

−

(a) 2 (b) 1 (c) -1 (d) 0

8. sin (45° + A) – cos (45° – A) = ?

(a) 1

2 (b) 1 (c) 0 (d)

3

2

9. cos (600 + A) + cos (60° - A) =?

(a) 1

2 (b) sin A (c) cos A (d) 3

10. 7 7

sin cos cos sin12 4 12 4

− =?

(a) 1 (b) 1

2 (c)

3

2 (d) 0

11. cos2 x4

+

– sin2 x

4

−

= ?

(a) cos2x

2 (b) cos 2x (c)

3

2sin 2x (d) 0

12. cosA + cos (120° - A) + cos (120° + A) =?

(a) 0 (b) 2cos A (c) - 1 (d) 1

13. cot 4

A

+

cot 4

A

−

=?

(a) - 1 (b) 1 (c) 0 (d) cot A

14. 2 sin (x – 45°) = ?

(a) sin x + cos x (b) sin x - cosx

(c) 2 (sin x – cos x) (d) cos x - sin x

15. 2 cos 4

x

+

=?

(a) sin x + cos x (b) sin x - cosx

(c) 2 (sin x – cos x) (d) cos x - sin x

16. 1 tan tan

1 tan tan

A B

A B

−

+= ?

(a) ( )( )

cos

cos

A B

A B

+

− (b) ( )

( )

cos

cos

A B

A B

−

+

(c) ( )( )

tan

tan

A B

A B

+

− (d) ( )

( )

tan

tan

A B

A B

−

+

17. =−

+

17sin17cos

17sin17cos

(a) 62tan (b) 56tan (c) 54tan (d) 73tan

18. =−

+

9sin9cos

9sin9cos

(a) cot 360 (b) 36tan (c) 18tan (d) None of these

Exercise - 1



65

19. =− 75cot75tan

(a) 32 (b) 32 + (c) 32 − (d) None of these

20. If ,sinsincoscos)cos( BABABA +=+ then ),( =

(a) (– 1, – 1) (b) (– 1, 1) (c) (1, – 1) (d) (1, 1)

21. tan 16° + tan 29° + tan 16° tan 29° = ?

(a) 1 (b) -1 (c) 0 (d) 2

22. ( )( )

− +

+ +

tan2 tan

1 tan2 tan =?

(a) tan ( + ) (b) –tan (2 – )

(c) tan (2 – ) (d) tan ( – )

23. If 5

3)cos( =− BA and ,2tantan =BA then

(a) 1

cos cos5

A B = −

(b) 2

sin sin5

A B =

(c) 5

1coscos −=BA

(d) 5

1sinsin −=BA

24. =

+

+

−

147cos

147sin

12sin12cos

12sin12cos

(a) 1 (b) – 1 (c) 0 (d) None of these

25. tan3 tan2 tanA A A− − =?

(a) AAA tan2tan3tan

(b) AAA tan2tan3tan−

(c) AAAAAA tan3tan3tan2tan2tantan −−

(d) None of these


66

1. 3

tan tan4 4

+ +

=?

(a) 1 (b) -1 (c) 2tan (d)

2. If A + B = 45° and tan A = m + 1 then find the value of

tan B in terms of m.

(a) m – 1 (b) m

m 1+ (c)

m

m 2

−

+ (d) 1 – m2

3. Find sin(A + B), if sinA = 3

5 and cos B =

5

13

(a) 48

65 (b)

63

65 (c)

12

13 (d) 1

4. Find sin (A -B) if cosA = 12

13 and cosB =

8

17

(a) 40

121− (b)

180

221− (c) −

140

221 (d)

180

221

5. cos (A - B) cos(A - B) =?

(a) cos2 A - sin2 B (b) cos2 B - sin2 A

(c) sin2 A - sin2 B (d) cos2 A - cos2 B

6. ( ) ( )( ) ( )

sin sin sin

cos cos cos

A B A A B

A B A A B

− + + +

− + + + =?

(a) sin (2A + B) (b) tan A

(c) cot B (d) tan B

7.

− + + tan cot

4 4A A =?

(a) 1 (b) 2 tan A (c) 2 cot A (d) 0

8. sin(A - B) sin(A + B) + sin (B - C) sin (B + C) + sin (C - A)

sin (C + A) = 0

(a) sin2 A – sin2 B – sin2C (b) sin2 A sin2 B sin2C

(c) 1 (d) 0

9. If 10

1sin =A and ,

5

1sin =B where A and B are

positive acute angles, then =+ BA

(a) (b) 2/ (c) 3/ (d) 4/

10. (1+tan 1) (1 + tan 2) (1 + tan3) …… (1 + tan 45°) = ?

(a) 221 (b) 222 (c) 223 (d) 224

11. (1-cot6°) (1-cot7°) (1 – cot7°) (1 – cot8°) …. (1 –cot39°)

= 2x, x = ?

(a) 15 (b) 16 (c) 17 (d) 18

12. If 7

1cos =P and ,

14

13cos =Q where P and Q both are acute

angles. Then the value of QP − is

(a) o30 (b) o60 (c) o45 (d) o75

13. sin163 cos347 sin73 sin167o o o o+ =?

(a) 0 (b) 1/2 (c) 1 (d) None of these

14. The value of + 150sin120cos330cos600sin is

(a) – 1 (b) 1 (c) 2

1 (d)

2

3

15. tan15 tan20 tan25 tan15 tan20 tan25

1 tan15 tan20 tan20 tan25 tan25 tan15

+ + −

− − − = ?

(a) 1 (b) 3 (c) 3− (d) 1

3

16. If tan ( + ) = 3

4 and tan ( – ) =

5

12 then find the

value of tan 2.

(a) 16

33 (b)

16

63 (c)

56

63 (d)

16

33

17. ( )sin

cos cos

A B

A B

−+

( )sin

cos cos

B C

B C

− +

( )sin

cos cos

C A

C A

− =?

(a) 2(tan A + tan B + tan C) (b) 0

(c) 1 (d) None of these

18. 2 ( cos 105° + sin 105°) = ?

(a) 0 (b) -1 (c) 1 (d) 3 1

2 2

−

19. If ,225=+ BA then( ) ( )

cot cot

1 cot 1 cot=

+ +

A B

A B?

(a) 1 (b) – 1 (c) 0 (d) ½

Exercise - 2



67

20. If tan,tan are the roots of the equation

),0( 02 =++ pqpxx then

(a) p−=+ )sin( (b) 1

)tan(−

=+q

p

(c) q−=+ 1)cos( (d) None of these

21.

+ + − tan tan

4 4A A =?

(a) cos 2 A (b) sec 2A (c) 2 sec 2A (d) 2 cos 2A

22. If ,cottan2tan BBA += then =− )tan(2 BA

(a) Btan (b) Btan2 (c) Bcot (d) Bcot2

23. =

−−

+

6sin

6cos 22

(a) 2cos2

1 (b) 0 (c) 2cos

2

1− (d)

2

1

24. If cot (A + B) = 4

3 and cot (A – B) =

12

5 then find the

value of tan 2B

(a) 16

63 (b)

56

63 (c)

16

33 (d)

56

33

25. sin (n + 1)A sin (n + 2)A + cos (n + 1)A cos (n + 2)A = ?

(a) cos (2n + 3)A (b) cos A

(c) –cos A (d) –cos (2n + 3)

26. if 2

1tan −=A and ,

3

1tan −=B then =+ BA

(a) 4

(b)

4

3 (c)

4

5 (d) None of these

27. If 5

4sin =A and ,

13

12cos −=B where A and B lie in first and

third quadrant respectively, then =+ )cos( BA

(a) 65

56 (b)

65

56− (c)

65

16 (d) –

65

16


68

Answer key

1 a 2 c 3 b 4 b 5 d

6 a 7 d 8 c 9 c 10 c

11 d 12 a 13 b 14 b 15 d

16 a 17 a 18 a 19 a 20 c

21 a 22 d 23 a 24 c 25 a

1. (a)

tan (45° - A) = tan45 tan

1 tan45 tan

A

A

+

− =

1 tan

1 1 tan

A

A

+

−

= 1 tan

1 tan

A

A

+

−

2. (c)

cot (45°-A) = cot cot 45 1

cot cot 45

A

A

+

−

= cot 1 1

cot 1

A

A

+

− =

cot 1

cot 1

A

A

+

−

3. (b)

( )

( )

tan tan

1 tan tan

A B B

A B B

+ −

+ +

= tan [(A + B) – B] = tan [A + B - B] = tan A

4. (b)

tan30 tan15

tan30 tan15 1

+

− =

+ −

−

tan30 tan15

1 tan30 tan15

= - tan (30°+ 15°) = - tan 45° = -1

5. (d)

tan A = 1

2 and tan B =

1

3

now tan (A + B) = tan tan

1 tan tan

A B

A B

+

−

=

1 1

2 31 1

12 3

+

−

=

5

65

6

= 1

6. (a)

sin22 cos22

sin22 cos22

+

− =

tan22 1

tan22 1

+

−

= 1 tan22

1 tan22

+ −

− = -tan (45° + 22°) = -tan 67°

7. (d)

tan69 tan66

1 tan69 tan66

+

− + 1

= tan (69° + 66°) + 1 = tan135° + 1 = -1 + 1 = 0

8. (c)

(sin 45° cos A + cos 45° sin A) – (cos 45° cos A + sin 45°

sin A)

= 1 1 1 1

cosA sin A cosA sin A2 2 2 2

+ − − = 0

9. (c)

cos(60°+A) + cos(60°-A)

= (cos60° cosA – sin 60° sin A) + (cos60° cosA + sin60°

sin A)

= 2 cos 60° cos A = 1

22

cosA = cos A

10. (c)

7 7

sin cos cos .sin12 4 12 4

−

= 7sin

12 4

−

=

3sin

3 2

=

11. (d)

Using cos (A + B)cos (A – B) = cos2 A – sin2 B

= + + − + − +

cos x x cos x x4 4 4 4

=

cos cos2x2

= 0

12. (a)

cosA + (cos120° cosA + sin 1200 sin A) + (cos120° cosA

– sin 120° sin A)

= cos A + 2 cos120° cosA = cosA +

−

12

2 cosA

= cosA – cos A = 0

Solutions

Exercise - 1



69

13. (b)

cot cot4 4

A A

+ −

= cot cot 1 cot cot 1

4 4

cot cot cot cot4 4

A A

A A

− +

+ −

= − +

+ −

cot 1 cot 1

cot 1 cot 1

A A

A A = 1

14. (b)

2sin (x -45°)

= 2 (sinx cos45° - sin450cosx)

= 2

−

1 1sin cos

2 2x x

= 2

2(sinx - cosx) = sinx – cosx

15. (d)

cos 4

x

+

= 2 × cos cos sin sin4 4

x x

−

= 2 1 1

cos sin2 2

x x

−

= cosx – sinx

16. (a)

1 tan tan

1 tan tan

A B

A B

−

+ =

sin sin1 .

cos cossin sin

1 .cos cos

A B

A BA B

A B

−

+

=

cos cos sin sin

cos coscos cos sin sin

cos cos

A B A B

A BA B A B

A B

−

+=

( )( )

cos

cos

A B

A B

+

−

17. (a)

oo

oo

17sin17cos

17sin17cos

−

+

Divided by o17cos in numerator and denominator,

1 tan 17

1 tan 17

o

o

+=

− = tan 45 tan 17

1 tan 45 tan 17

o o

o o

+

−

= ( )tan 45 17+ = tan620

18. (a)

cos 9 sin 9

cos9 sin 9

o o

o o

+

− =

o

o

9tan1

9tan1

−

+

= ( )tan 45 9o o+ = tan54o = tan (900 - 360) = cot 360

19. (a)

tan75 cot75o o− = tan75 tan15o o−

32)32()32( =−−+=

20. (c)

BABABA sinsincoscos)(cos +=+

But BABABA sinsincoscos)(cos −=+ 1, 1 = = −

21. (a)

tan (16° + 29°) = tan16 tan29

1 tan16 tan29

+

−

tan 45° = 1 = tan16 tan29

1 tan16 tan29

+

−

1 – tan16° tan29° = tan16° + tan29°

tan16° + tan29° + tan16° tan29° = 1

22. (d)

( )( )

tan2 tan

1 tan2 .tan

− +

+ + = tan (2 – – ) = tan ( – )

23. (a)

5

3)(cos =− BA

3sinsin5coscos5 =+ BABA …..(i)

From 2nd relation,

BABA coscos2sinsin = .....(ii)

5

1coscos =BA and 5sin sin 2A B =

24. (c)

o

o

oo

oo

147cos

147sin

12sin12cos

12sin12cos+

+

−

o

o

o

147tan12tan1

12tan1+

+

−=

( ) ( )tan 45 12 tan 180 33o o o o= − + −

= tan 330 – tan 330 = 0

25. (a)

( )tan tan 2

tan 3 tan 21 tan tan 2

A AA A A

A A

+= + =

−

tan 3 tan 3 tan 2 tan tan tan 2A A A A A A − = +

AAAAAA tan2tan3tantan2tan3tan =−−


70

Answer key

1 b 2 c 3 b 4 c 5 a

6 b 7 d 8 d 9 d 10 c

11 c 12 b 13 b 14 a 15 b

16 d 17 b 18 c 19 d 20 b

21 c 22 c 23 a 24 a 25 b

26 b 27 d

1. (b)

tan3

tan4 4

+ +

=

+

−

tan tan4

1 tan tan4

+

−

3tan tan

43

1 tan tan4

= 1 tan

1 1 tan

+

− ×

( )

− +

− −

1 tan

1 1 tan

= ( )( )

+

−

1 tan

1 tan×

( )( )

tan 1

1 tan

−

+ =

( )( )

tan 1

1 tan

−

−

=( )( )

−−

−

1 tan

1 tan = -1

2. (c)

tan (A + B) = tanA tanB

1 tanAtanB

+

−

1 = ( )

m 1 tanB

1 m 1 tanB

+ +

− +

m + 1 + tan B = 1 – m tan B – tan B

(2 + m) tan B = -m tan B = m

m 2

−

+ 3. (b)

sin A = 3

5

cosA = 21 sin A− =

23

15

−

=

16

25 =

4

5

and cos B = 5

13so

sinB = 21 cos B− =

25

113

−

=

144

169 =

12

13

Now sin(A + B) = sin A cosB + cos A sinB

=3 5 4 12

5 13 3 13 + =

15 48 15 48

65 65 65

++ = =

63

65

4. (c)

cos A = 12

13

sin A = 21 cos A− =

212

113

−

=

5

13

and cos B = 8

17

sin B = 21 cos B− =

28

117

−

=

15

17

Now, sin (A - B) = sin A cosB – cosA sin B

= 5 8 12 15

13 17 13 17 − = −

40 180

221 221= −

140

221

5. (a)

cos(A + B) cos(A-B)

= (cos A cos B – sin A sin A sin B) (cos A cos B + sin A

sinB)

= cos2 A cos2 B –sin2A sin2B

= cos2 Acos2B- (1- cos2A) (1- cos2B)

= cos2A cos2B- (1- cos2B-cos2A+cos2Acos2B)

= cos2Acos2B-1+cos2B + cos2 A – cos2Acos2B

= cos2A -1 + cos2 B = cos2 A – (1-cos2B)

= cos2 A - sin2 B

6. (b)

( ) ( )( ) ( )

sin sin sin

cos cos cos

A B A A B

A B A A B

− + + +

− + + +

( ) ( )( ) ( )

− + + +=

+ + + −

sin cos cos sin sin sin cos cos sin

cos cos sin sin cos cos cos sin sin

A B A B A A B A B

A B A B A A B A B

= sin 2sin cos

cos 2cos cos

A A B

A A B

+

+=

( )sin 1 2cos

cos (1 2cos )

A B

A B

+

+

= sin

cos

A

A = tan A

7. (d)

tan cot4 4

A A

− + +

= tan tan cot cot 1

4 4

1 tan tan cot cot4 4

A A

A A

− −

+

+ +

Exercise - 2



71

= − −

++ +

tan 1 cot 1

1 tan 1 cot

A A

A A =

−−

++

+

11

tan 1 tan11 tan

1tan

A AA

A

= − −

++ +

tan 1 1 tan

1 tan tan 1

A A

A A=

− −−

+ +

tan 1 tan 1

1 tan tan 1

A A

A A = 0

8. (d)

sin (A - B) sin (A + B) + sin (B - C) sin (B + C) + sin (C - A)

sin (C + A)

= sin2A – sin2B + sin2B – sin2C + sin2C – sin2 A

[ sin (A - B) sin (A + B) = sin2 A – sin2 B] = 0

9. (d)

BABABA sincoscossin)(sin +=+

10

11

5

1

5

11

10

1−+−=

1 4 1 9

5 1010 5= + =

1 5 1(2 3)

50 50 2+ = =

4

sin)(sin

=+ BA Hence, 4

=+ BA

10. (c)

(1+ tan1°) (1 + tan 2°) (1 + tan 3°) (1 +tan 4°) ……………

(1 + tan 45°)

= ( 1+ tan 1°) (1+ tan 44°) (1 +tan2°) (1 + tan 43°)

………22 times (1 + tan45°)

= 2 × 2 ………………… 22 times × (1 × 1)

= 222 × 2 = 223

11. (c)

( 1 – cot6°) (1-cot39°) (1- cot7°) (1 – cot38°)…… 34

terms

= [(1 – cot6°) (1 – cot39°)] [(1- cot7°) (1 – cot38°)]

……….. 17 terms

= [1- cot6° - cot39° + cot6° cot31°] [(1cot7° - cot39° +

cot7° cot39°)] ……………… 17terms

= 2 × 2 x ……. 17 times = 217

x = 17

12. (b)

14

13cos,

7

1cos == QP

QPQPQP sinsincoscos)cos( +=−

1 13 48 27

7 14 7 14= +

= 13 36 1

cos6098 2

o+= = oQP 60=−

13. (b)

oooo 167sin73sin347cos163sin +

)1790cos()13360cos()17180sin( oooooo −+−−=

)13180sin( oo −

sin17 cos13 cos17 sin13o o o o= +

( )sin 17 13 sin30 1/ 2o o= + = =

14. (a)

oooo 150sin120cos330cos600sin +

oooo 60cos30sin30cos60sin −−=

sin (60 30 )o o= − + = sin90 1− = −

15. (b)

tan15 tan20 tan25 tan15 tan20 tan25

1 tan15 tan20 tan20 tan25 tan25 tan15

+ + −

− − −

= tan (150 + 200 + 250) = tan 60° = 3

16. (d)

tan 2 = tan [( + ) + ( – )]

= ( ) ( )

( ) ( )

tan tan

1 tan tan

+ + −

− + − =

3 5

4 123 5

14 12

+

−

= 56

33

17. (b)

( )−sin

cos cos

A B

A B=

−sin cos cos sin

cos cos

A B A B

A B=tanA –tanB

( )−sin

cos cos

B C

B C=

−sin cos cos sin

cos cos

B C B C

B C= tanB-tanC

( )−sin

cos cos

C A

C A=

sin cos cos sin

cos cos

C A C A

C A

−= tanC-tanA

Now

( )sin

cos cos

A B

A B

−+

( )sin

cos cos

B C

B C

− +

( )sin

cos cos

C A

C A

−

= tan A – tan B + tan B –tan C + tan C –tan A = 0

18. (c)

( )2 cos105 sin105 +

= ( ) ( )2 cos 60 45 sin 60 45 + + +

= 2 [cos60°cos45°-sin60°sin45°+sin60°cos45° +

cos60°sin45°]

= 21 1 3 1 3 1 1 1

2 2 2 22 2 2 2

− + +


72

= 2 1 3 3 1

2 2 2 2 2 2 2 2

− + +

= 1 1

22 2 2 2

+

=

2 2

2 2 = 1

19. (d)

A + B = 2250 tan( ) tan 225+ = oA B

tan tan 1 tan tanA B A B + = −

(1+tanA)(1+tanB) = 2

1

1cot A

+

11

cot B

+

= 2

( ) ( )

cot cot

1 cot 1 cot+ +

A B

A B=

1

2

20. ( b)

Since tan,tan are the roots of the equation

2 0x px q+ + =

,tantan p−=+ q= tantan

tan tan

tan ( )1 tan tan

++ =

− ( )1 1

p p

q q

−= =

− − −

21. (c)

= tan4

A

+

+tan4

A

−

=+ −

+− +

1 tan 1 tan

1 tan 1 tan

A A

A A

( ) ( )

( )( )

+ + −

− +

2 21 tan 1 tan

1 tan 1 tan

A A

A A=

( )( )

+

−

2

2

2 1 tan

1 tan

A

A

=

+

−

2

2

2

2

sin2 1

cos

sin1

cos

A

A

A

A

= ( )+

−

2 2

2 2

2 cos sin

cos sin

A A

A A=

2 1

cos2A

= 2 sec

2A

22. (c)

,cottan2tan BBA +=

+

−=−

BA

BABA

tantan1

tantan2)(tan2

= (2 tan cot tan )

21 (2 tan cot ) tan

B B B

B B B

+ −

+ +

= 2

tan cot2

2(1 tan )

B B

B

+

+ B

B

BBcot

)tan1(

)1(tancot2

2

=+

+=

23. (a)

−−

+

6sin

6cos 22

+−+

−++=

66cos

66cos

)]cos()cos(sincos[ 22 BABABA −+=−

1

cos cos2 cos23 2

= =

24. (a)

tan 2B = ( ) ( )

1 1

cot2B cot A B A B=

+ − −

= ( ) ( )

( ) ( )

cot A B cot A B

cot A B cot A B 1

− − +

− + + =

12 4 16165 3 15

12 4 63 631

5 3 15

−

= =

+

25. (b)

cos x cos y + sin x sin y = cos (x – y)

sin (n + 1)A sin (n + 2)A + cos (n + 1)A cos (n + 2)A

= cos {(n + 1)A – (n + 2)A}

= cos(n+1–n–2)A = cos (-A) = cos A

26. (b)

2

1tan −=A and

3

1tan −=B

1 1tan tan 2 3tan( ) 1

1 11 tan tan1

2 3

A BA B

A B

− −+

+ = = = −−

−

3

tan ( ) tan4

A B

+ =

Hence, 3

4A B

+ =

27. (d)

5

4sin =A and

13

12cos −=B

BABABA sinsincoscos)(cos −=+

169

1441

5

4

13

12

25

161 −−

−−=

65

16

13

5

5

4

13

12

5

3−=

−−−=

(Since A lies in first quadrant and B lies in third

quadrant)

Rose

James Raja

007

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Trigonometry

Transformation of a Product

into a sum or Difference

Chapter - 07



73

TRANSFORMATION OF A

PRODUCT INTO A SUM OR DIFFERENCE

(i) 2 sin A cos B = sin (A + B) + sin (A – B)

(ii) 2 cos A sin B = sin (A + B) – sin (A – B)

(iii) 2 cos A cos B = cos (A + B) + cos (A – B)

(iv) 2 sin A sin B = cos (A – B) – cos (A + B)

(i) sin (A + B) + sin (A – B)

= (sin A cos B + cos A sin B) + (sin A cos B – cos A sin B)

= 2 sin A cos B

2 sin A cos B = sin (A + B) + sin (A – B)

(ii) sin (A + B) – sin (A – B)

= (sin A cos B + cos A sin B) – (sin A cos B – cos A sin B)

= 2 cos A sin B

2 cos A sin B = sin (A + B) – sin (A – B)

(iii) cos (A + B) + cos (A – B)

= (cos A cos B – sin A sin B) + (cos A cos B + sin A sin B)

= 2 cos A cos B

2 cos A cos B = cos (A + B) + cos (A – B)

(iv) cos (A – B) – cos (A + B)

= cos A cos B + sin A sin B) – (cos A cos B – sin A sin B)

= 2 sin A sin B

2 sin A sin B = cos (A – B) – cos (A + B)

TRANSFORMATION OF A

SUM OR DIFFERENCE INTO A PRODUCT

(i) sin C + sin D = 2 sin C D C D

cos2 2

+ −

(ii) sin C – sin D = 2 cos C D C D

sin2 2

+ −

(iii) cos C + cos D = 2 cos C D C D

cos2 2

+ −

(iv) cos C – cos D = 2 sin C D D C

sin2 2

+ −

(i) sin C + sin D

= sin (A + B) + sin (A – B)

= (sin A cos B + cos A sin B) + (sin A cos B – cos A sin B)

= 2 sin A cos B = 2 sin C D C D

cos2 2

+ −

sin C + sin D = 2 sin C D C D

cos2 2

+ −

(ii) sin C – sin D

= sin (A + B) – sin (A – B)

= (sin A cos B + cos A sin B) – (sin A cos B – cos A sin B)

= 2 cos A sin B = 2 cos C D C D

sin2 2

+ −

sin C – sin D = 2 cos C D C D

sin2 2

+ −

(iii) cos C + cos D

= cos (A + B) + cos (A – B)

= (cos A cos B – sin A sin B) + (cos A cos B + sin A sin B)

= 2 cos A cos B = 2 cos C D C D

cos2 2

+ −

cos C + cos D = 2 cos C D C D

cos2 2

+ −

(iv) cos C – cos D

= cos (A + B) – cos (A – B)

= (cos A cos B – sin A sin B) – (cos A cos B + sin A sin B)

= – 2 sin A sin B = 2 sin A (– sin B)

= 2 sin A sin (– B) [ – sin x = sin (– x)]

= 2sinC D C D

sin2 2

+ − −

Using 2

= 2 sinC D D C

sin2 2

+ −

cos C – cos D = 2 sinC D D C

sin2 2

+ −

Transformation of a Product into a Sum or

Difference 7


74

Express the following as a sum or difference:

1. 2 sin 5x cos 3x

Sol. 2 sin 5x cos 3x = sin (5x + 3x) + sin (5x – 3x)

= sin 8x + sin2x

2. 7x 5x

2cos sin2 2

Sol. 7x 5x

2cos sin2 2

= + − −

7x 5x 7x 5xsin sin

2 2 2 2

[ 2 cos A sin B = sin (A + B) – sin (A – B)]

= sin 6x – sinx

3. x 3x

sin sin2 2

Sol. x 3x 1 x 3xsin sin 2sin sin

2 2 2 2 2

=

= 1 x 3x x 3x

cos cos2 2 2 2 2

− − +

[ 2 sin A sin B = cos (A – B) – cos (A + B)

= 1

2 [cos (–x) – cos 2x] =

1

2 [cos x – cos 2x]

[ cos (– x) = cos x]

Ex.4. cosec (45° + x) cosec (45° – x) =?

Sol. cosec (45° + x) cosec (45° – x)

= ( ) ( )

2

2sin 45 x sin 45 x + −

= ( ) ( ) ( ) ( )

2

cos 45 x 45 x cos 45 x 45 x + − − − + + −

[ 2 sin A sin B = cos (A – B) – cos (A + B)]

= 2 2

cos2x cos90 cos2x 0=

− − = 2 sec 2x

5. cos 5x – cos 11x =?

Sol. cos 5x – cos 11x

= 2 sin 5x 11x 11x 5x

sin2 2

+ −

C D D CcosC cosD 2sin sin

2 2

+ − − =

= 2 sin 8x sin 3x

6. sin 80° - sin 20°=?

Sol. sin 80° - sin 20°

= 2cos 80 20 80 20

sin2 2

+ −

C D C DsinC sinD 2cos sin

2 2

+ − − =

= 2 cos 50° sin 30° = 2 cos 50°1

2 = cos 50°

7. sin3A sin5A

cos3A cos5A

+

+ =?

Sol. sin3A sin5A

cos3A cos5A

+

+

=

+ −

+ −

5A 3A 5A 3A2sin cos

2 2

5A 3A 5A 3A2cos cos

2 2

= 2sin4AcosA

2cos4AcosA= tan4A

8. sin 65° + cos 65° = ?

Sol. sin 65° + cos 65° = sin 65° + cos (90° – 25°)

= sin 65° + sin 25°

= 65 25 65 252sin cos

2 2

+ −

= 2 sin 45°cos 20° = 2 × 1

2 × cos 20° = 2 cos20

9. +

+

sinA sinB

cosA cosB=?

Sol. +

+

sinA sinB

cosA cosB=

+ −

+ −

A B A B2sin cos

2 2A B A B

2cos cos2 2

= A B

tan2

+

10. cot x (sin 5x – sin 3x)

Sol. cot x (sin 5x – sin 3x)

= cot x 5x 3x 5x 3x

2cos sin2 2

+ −

A B A BsinA sinB 2cos sin

2 2

+ − − =

Examples



75

= cosx

sinx × 2 cos 4x. sin x = 2 cos 4x cos x

10. sin 10° + sin 20° + sin 40° + sin 50° = ?

Sol. (sin 50° + sin 10°) + (sin 40° + sin 20°)

= 50 10 50 102sin cos

2 2

+ −

40 20 40 202sin cos

2 2

+ − +

= 2 sin 30° cos 20° + 2 sin 30° cos 10°

= 2 × 1

2 × cos 20° + 2 ×

1

2 × cos 10°

= cos 20° + cos 10°

11. cos 20° cos 40° cos 60° cos 80° =?

Sol. cos 60° cos 80° cos 40° cos 20°

= 1

2 cos 80° cos 40° cos 20°

1cos60

2

=

= 1

4 (2 cos 80° cos 40°) cos 20°

= 1

4 [cos (80° + 40°) + cos (80° – 40°)] cos 20°

= 1

4 [cos 120° + cos 40°] cos 20°

= 1 1cos40 cos20

4 2

− +

= 1 1cos20 cos40 cos20

4 2

− +

= 1

8− cos 20° +

1

8 [2 cos 40° cos 20°]

= 1

8− cos 20° +

1

8 [cos (40° + 20°) + cos (40° – 20°)]

= 1

8− cos 20° +

1

8 [cos 60° + cos 20°]

= 1 1 1cos20 cos20

8 8 2

− + +

= 1

8− cos 20° +

1 1 1cos20

16 8 16+ =

12. 4 cos 12° cos 48° cos 72° =?

Sol. L.H.S. = 4 cos 12° cos 48° cos 72°

= 2 (2 cos 48° cos 12°) cos 72°

= 2 [cos (48° + 12°) + cos (48° – 12°)] cos 72°

[ 2 cos A cos B = cos (A + B) + cos (A – B)]

= 2 (cos 60° + cos 36°) cos 72°

= 2 cos 60° cos 72° + 2 cos 36° cos 72°

= 2 × 1

2 cos 72° + 2 cos 72° cos 36°

= cos 72° + [cos (72° + 36°) + cos (72° – 36°)]

= cos 72° + cos 108° + cos 36°

= cos 72° + cos (108° – 72°) + cos 36°

= cos 72° – cos 72° + cos 36°

[ cos (180° – x) = – cos x] = cos 36°


76



77

1. 5

cos cos12 12

=?

(a) 1 (b) 1

4 (c) 1/2 (d)

3

2

2. 2 sin 45° sin 15° =?

(a) 1 (b) 3 1

2

+ (c) 1/2 (d)

3 1

2

−

3. sin 14x + sin 2x =?

(a) 2 sin 8x cos 6x (b) 2 sin 6x cos 8x

(c) 1/2 (d) sin 4x cos 10x

4. 1

2 (cos 7x + cos x)=?

(a) cos 4x cos 3x (b) sin 4x sin 3x

(c) cos 4x sin 3x (d) 1

5. sin 47° + cos 77° = ?

(a) sin 17° (b) cos 17°

(c) 2 sin 73° (d) 2cos73°

6. sin 25° cos 115° = ?

(a) 1

2 (cos 40° – 1) (b)

1

2 (cos 50° – 1)

(c) 1

2 (sin 40° – 1) (d) 1

7. cos 80° + cos 40° – cos 20° = ?

(a) 1 (b) 0 (c) 1/2 (d) 2

8. sin50 sin70 sin10− + =?

(a) 1 (b) 1/2 (c) 0 (d) 2

9. sin7A sin5A

cos7A cos5A

−

+ =?

(a) cot 6A (b) cot A (c) tan 6A (d) tan A

10. −

−

cos9x cos5x

sin17x sin3x=?

(a) sin2x

cos10x

− (b)

cos2x

sin10x

− (c) tan 7x (d) cot 7x

11. cot 4x (sin 5x + sin 3x)=?

(a) 2 sin 4x cos x (b) 2 cos 4x sin x

(c) 2 cos 4x cos x (d) 2

12. The value of ++ 172cos68cos52cos is

(a) 0 (b) 1 (c) 2 (d) 2

3

13. cos 55° + cos 65° + cos 175° = ?

(a) 1 (b) 0 (c) 1/2 (d) 2

14. cos7A cos11A

cos7A cos11A

+

− =?

(a) 1 (b) cot 9

tan 2A

A (c)

tan9

tan 2A

A (d)

tan9

cot 2A

A

15. cos 8 – cos 3 =

(a)

−11 5

2sin sin2 2

(b) 13 3

2sin sin2 2

−

(c) 11 5

2cos cos2 2

− (d)

11 52sin cos

2 2

16. 5 2

cos sin7 7

=?

(a) 1 3

cos2 7

(b) 0 (c) -

1 3sin

2 7

(d)-1

17. cos 35°cos 25°=?

(a) 1

sin102

(b) 1

cos102

(c) 1/2 (d) 1 1

cos104 2

+

18. cos 15° – sin 15° = ?

(a) 3

2 (b) 3

2 2 (c) 1 (d)

1

2

19. If ,)sin(

)sin(

ba

ba

yx

yx

−

+=

−

+then

y

x

tan

tan is equal to

(a) a

b (b) b

a (c) ab (d) None of these

20. If cos (A – B) = 3 cos (A + B) then find the value of tan A

tan B

Exercise - 1


78

(a) 1 (b) 2 (c) 1

2 (d)

1

2−

21. =+++

+++

9cos7cos5cos3cos

9sin7sin5sin3sin

(a) 3tan (b) 3cot (c) 6tan (d) 6cot

22. 9 3 5

2cos cos cos cos13 13 13 13

+ + =?

(a) – 1 (b) 0 (c) 1 (d) None of these

23. If ,cotcoscos

sinsinB

AC

CA=

−

− then A,B,C are in

(a) A.P. (b) G.P.

(c) H.P. (d) None of these

24. If ,tancostan = then =2

tan2

(a) )sin(

)sin(

−

+ (b)

)cos(

)cos(

+

−

(c) )sin(

)sin(

+

− (d)

)cos(

)cos(

−

+

25. If ,cos21

=+x

x then =+3

3 1

xx

(a) 3cos (b) 3cos2

(c) 3cos2

1 (d) 3cos3

1



79

1. cot 4x cos3x cos2x

sin4x sin3x sin2x

+ +

+ + =?

(a) cot 3x (b) cot2x (c) tan3x (d) tan2x

2. sin 10° + sin 20° + sin 40° + sin 50° =?

(a) 2 sin 15° sin 5° (b) 2 sin 15° cos 5°

(c) 2 cos 15° sin 5° (d) 2 cos 15° cos 5°

3. =++−

−++

)cos()sin(

)cos()sin(

ABAB

ABAB

(a) BB

BB

sincos

sincos

−

+ (b) AA

AA

sincos

sincos

−

+

(c) AA

AA

sincos

sincos

+

− (d) None of these

4. 9 3 5


+ + =?

(a) 3 (b) 2 (c) 0 (d) 1

5. cos2 A + cos2 B – 2 cos A cos B cos (A + B) =?

(a) sin2 (A + B) (b) cos2 (A+B)

(c) sin2 (A - B) (d) cos2 (A - B)

6. ( ) ( )( ) ( ) − − −

− − −

2sin cos sin 2

2sin cos sin 2=?

(a)

sin

sin (b)

cos

cos (c)

cos

cos (d)

sin

sin

7. cos 7x + cos 5x + cos 3x + cos x =?

(a) 4 cos x cos 2x cos 4x

(b) 2 cos x cos 2x cos 4x

(c) cos x cos 2x cos 4x

(d) 4 cos x cos 3x cos 5x

8. If cosec A + sec A = cosec B + Sec B then find the value of

cotA B

2

+

(a) tan A tan B (b) cot A cot B

(c) tan A cot B (d) cot A tan B

9. If three angles A, B and C are in A.P. then find the value

of sin A sinC

cosC cosA

−

−.

(a) cot (A + C) (b) cot B

(c) cot A (d) tan (B + C)

10. If ),2sin(sin += ab then =−

+

ba

ba

(a) )tan(

tan

+ (b)

)cot(

cot

−

(c) )cot(

cot

+

− (d)

)cot(

cot

+

11. cos 2x cos x

2 – cos 3x cos

9x

2 =?

(a) 5x

sin5xsin2

(b) cos 5x cos5

2

x

(c) sin 5x cos5

2

x (d) cos 5x sin

5

2

x

12. If sin3sin2sinsin =++

and cos3cos2coscos =++ , then is equal to

(a) 2/ (b) (c) 2 (d) 6/

13. =−− xxx 5cos3coscos2

(a) xx 23 sincos16 (b) xx 23 cossin16

(c) xx 23 sincos4 (d) xx 23 cossin4

14. If ,coscos BmA = then

(a) 2

tan1

1

2cot

AB

m

mBA −

−

+=

+

(b) 2

cot1

1

2tan

AB

m

mBA −

−

+=

+

(c) 2

tan1

1

2cot

BA

m

mBA −

−

+=

+

(d) None of these

15. If sin A = sin B and cos A = cos B then which of the

following is true

(a) A Bcos

2

+

= 0 (b) A Bsin

2

−

= 0

(c) A Bsin

2

+

= 0 (d) None is true

16. If 0coscoscos =++ yx and ,0sinsinsin =++ yx then

Exercise - 2


80

=

+

2cot

yx?

(a) sin (b) cos

(c) cot (d)

+

2sin

yx

17. The value of oo 70cos470cot + is

(a) 3

1 (b) 3 (c) 32 (d)

2

1

18. sin8xcosx sin6xcos3x

cos2xcosx sin3xsin4x

−

− =?

(a) tan3x (b) cot2x (c) sec2x (d) tan 2x

19. cos6x 6cos4x 15cos2x 10

cos5x 5cos3x 10cosx

+ + +

+ + =?

(a) sin2x (b) cos3x (c) 3sin2x (d) 2cos x

20. If ,coscos,sinsin DBACBA =+=+ then the value of

=+ )sin( BA

(a) CD (b) 22 DC

CD

+

(c) CD

DC

2

22 + (d) 22

2

DC

CD

+

21. =+−− 81tan63tan27tan9tan

(a) 1/2 (b) 2 (c) 4 (d) 8

22. If tan1 = k cot 2 then ( )( )

1 2

1 2

cos

cos

−

+ =

(a) 1 k

1 k

+

− (b)

1 k

1 k

−

− (c)

k 1

k 1

+

− (d)

k 1

k 1

−

+

23. Find the value

+ +

+ +

cos3 2cos5 cos7

cos 2cos3 cos5 + sin 2 tan 3

(a) cos 2 (b) tan 2 (c) cos 4 (d) sec 2

24. + + +

+ + +

sin sin2 sin4 sin5

cos cos2 cos4 cos5

A A A A

A A A A =?

(a) tan8A (b) cotA (c) tan6A (d) tan3A

25. In quadrilateral ABCD if sin 2

A B+

cos2

A B−

sin cos 22 2

C D C D+ − + =

, then value of sin

2

Asin

2

Bsin

2

Csin

2

D

(a) 0 (b) 1

16 (c)

1

4 (d) 1

26. 2 sin (60° – x) cos (30° + x)

(a) 1 - +cos2x 3 sin2x

2 2 (b) 1 -

cos2x 3 sin2x

2 2−

(c) 0 (d) 1 + cos2x 3 sin2x

2 2−



81

Answer key

1 b 2 d 3 a 4 a 5 b

6 c 7 b 8 c 9 d 10 a

11 c 12 a 13 b 14 b 15 a

16 c 17 d 18 b 19 c 20 c

21 c 22 b 23 a 24 c 25 b

1. (b)

5 1 5cos cos 2cos cos

12 12 2 12 12

=

= 1 5 5

cos cos2 12 12 12 12

+ + −

= 1

cos cos2 2 3

+

= 1 1

02 2

+

=

1

4

2. (d)

2 sin 45° sin 15°

= cos (45° – 15°) – cos (45° + 15°)

= cos 30° - cos 60° = 3 1 3 1

2 2 2

−− =

3. (a)

sin 14x + sin 2x

= 2 sin14x 2x 14x 2x

cos2 2

+ −

C D C DsinC sinD 2sin cos

2 2

+ − + =

= 2 sin 8x cos 6x

4. (a)

1

2 (cos 7x + cos x) = 1 7x x 7x x

2cos cos2 2 2

+ −

C D C DcosC cosD 2cos cos

2 2

+ − + =

= cos 4x cos 3x

5. (b)

sin 47° + cos 77°

= sin 47° + cos 13° = 2 sin 30°cos 17° = cos 17°

6. (c)

sin 25° cos 115°

= 1

2 (2 sin 25° cos 115°) =

1

2 (sin 140° – sin 90°)

= 1

2 [sin (180° – 40°) – 1] =

1

2 (sin 40° – 1)

7. (b)

cos 80° + cos 40° – cos 20°

= 2 cos 60°.cos 20° – cos 20°

= cos 20° – cos 20° = 0

8. (c)

ooo 10sin70sin50sin +−

2cos60 sin10 sin10o o o= − + sin10 (1 2 cos 60 ) 0o o= − =

9. (d)

sin7A sin5A

cos7A cos5A

−

+ =

A 5A A 5A2sin cos

2 2

A 5A A 5A2cos cos

2 2

− +

+ −

= 2sinAcos6A

2cos6AcosA = tan A

10. (a)

−

−

cos9x cos5x

sin17x sin3x

=

+ − −

+ −

9x 5x 9x 5x2sin sin

2 2

17x 3x 17x 3x2cos sin

2 2

= 2sin7xsin2x sin2x

2cos10xsin7x cos10x

− −=

11. (c)

cot 4x (sin 5x + sin 3x)

Solutions

Exercise - 1


82

= cot 4x 5x 3x 5x 3x

2sin cos2 2

+ −

A B A BsinA sinB 2sin cos

2 2

+ − + =

= cos4x

sin4x × 2 sin 4x cos x = 2 cos 4x cos x

12. (a)

ooo 172cos68cos52cos ++

ooo 68cos)172cos52(cos ++=

ooo 68cos60cos112cos2 +=

cos 112 cos 68o o= + 2 cos (90 ) cos 22 0o o= =

13. (b)

cos 55° + cos 65° + cos 175°

= 55 65 65 552cos cos cos175

2 2

+ − +

= 2 cos 60°.cos 5° + cos 175°

= 2 × 1

2 × cos 5° + cos (180° – 5°) = cos 5° – cos 5° = 0

14. (b)

cos7A cos11A

cos7A cos11A

+

−=

+ −

+ −

7A 11A 11A 7A2cos cos

2 2

7A 11A 11A 7A2sin sin

2 2

= cos9Acos2A

sin9Asin2A = cot 9A cot 2A =

cot 9

tan 2A

A

15. (a)

cos 8 – cos 3

= + −

8 3 3 82sin sin

2 2

= −

11 52sin sin

2 2 =

−

11 52sin sin

2 2

16. (c)

5 2

cos sin7 7

=

1 5 22cos sin

2 7 7

= 1 5 2 5 2

sin sin2 7 7 7 7

+ − −

= 1 3sin sin

2 7

−

= -1 3

sin2 7

17. (d)

1

2 (cos 35°cos 25°)

= 1

2 [cos (35° + 25°) + cos (35° – 25°)]

= 1

2 (cos 60° + cos 10°)

= +

1 1cos10

2 2 =

1 1cos10

4 2+

18. (d)

cos 15° – sin 15° = sin 75° – sin 15°

= 2cos75 15

2

+ sin

75 15

2

−

= 2cos 45° sin 30° = =1 1 1

222 2

19. (b)

ba

ba

yx

yx

−

+=

−

+

)(sin

)(sin

)()(

)()(

)(sin)(sin

)(sin)(sin

baba

baba

yxyx

yxyx

−−+

−++=

−−+

−++

2sin cos 2

2cos sin 2

x y a

x y b =

tan

tan

x a

y b =

20. (c)

( )( )

cos A B

cos A B

−

+ = 3

Applying C & D

( ) ( )( ) ( )

cos A B cos A B 3 1

3 1cos A B cos A B

− + + +=

−− − +

2cosAcosB

2sinAsinB = 2 tan A tan B =

1

2

21. (c)

9cos7cos5cos3cos

9sin7sin5sin3sin

+++

+++

)7cos5(cos)9cos3(cos

)7sin5(sin)9sin3(sin

+++

+++=

cos6cos23cos6cos2

cos6sin23cos6sin2

+

+=

2sin 6 (cos3 cos )

2cos6 (cos3 cos )

+=

+

= tan6

22. (b)

9 3 5


+ +



83

9 4

2cos cos 2cos cos13 13 13 13

= +

+=

13

4cos

13

9cos

13cos2

5

2cos 2cos cos 013 2 26

= =

,

= 0

2cos

23. (a)

BAC

CAcot

coscos

sinsin=

−

− BCACA

CACA

cot

2sin

2sin2

2sin

2cos2

=−+

−+

BCA

cot2

)(cot =

+

2

CAB

+=

Thus A, B, C will be in A.P.

24. (c)

2 1 costan

2 1 cos

−=

+ =

tan tan

tan tan

−

+=

sin( )

sin( )

−

+

25. (b)

cos21

=+x

x

+−

+=+

xx

xx

xx

xx

113

113

3

3

= 3(2cos ) 3(2cos ) − = 38cos 6cos −

= 32(4cos 3cos ) − = 2cos3

Alternate:

Put 1=x = 0

Then 3cos2213

3 ==+x

x

Answer key

1 a 2 d 3 b 4 c 5 a

6 d 7 a 8 a 9 b 10 c

11 c 12 a 13 a 14 a 15 b

16 c 17 b 18 d 19 d 20 d

21 c 22 a 23 a 24 d 25 c

26 d

1. (a)

cot 4x cos3x cos2x

sin4x sin3x sin2x

+ +

+ +

= ( )( )

cos4x cos2x cos3x

sin4x sin2x sin3x

+ +

+ +

=

6x 2x2cos .cos cos3x

2 2

6x 2x2cos .cos sin3x

2 2

+

+

= 2cos3xcosx cos3x

2sin3xcosx sin3x

+

+ =

( )( )

cos3x 2cosx 1

sin3x 2cosx 1

+

+ = cot 3x

2. (d)

sin 10° + sin 20° + sin 40° + sin 50°

= (sin 50° + sin 10°) + (sin 40° + sin 20°)

= 50 10 50 102sin cos

2 2

+ −

40 20 40 202sin cos

2 2

+ − +

= 2 sin 30° cos 20° + 2 sin 30° cos 10°

= 2 sin 30° [cos 20° + cos 10°]

= 2 sin 30°20 10 20 10

2cos cos2 2

+ −

= 2 × 1

2 [2 cos 15° cos 5°] = 2 cos 15° cos 5°

3. (b)

)(cos)(sin

)(cos)(sin

ABAB

ABAB

++−

−++

( )( )

sin ( ) sin[90 ]

sin ( ) sin[90 ]

o

o

B A B A

B A A B

+ + − −=

− + − −

)45(cos)45(sin2

)45(cos)45(sin2

BA

BAoo

oo

−−

−+=

Exercise - 2


84

sin ( 45 )

sin (45 )

o

o

A

A

+=

− =

cos sin

cos sin

A A

A A

+

−

4. (c)

9 3 5


+ +

= 9 9 3 5cos cos cos cos

13 13 13 13 13 13

+ + − + +

[ 2 cos A cos B = cos (A + B) + cos (A – B)]

= 10 8 3 5

cos cos cos cos13 13 13 13

+ + +

= 3 5 3 5cos cos cos cos

13 13 13 13

− + − + +

= 3 5 3 5


− − + +

[ cos ( – x) = – cos x] = 0

5. (a)

cos2 A + cos2 B – 2 cos A cos B cos (A + B)

= cos2 A + cos2 B – [cos (A + B) + cos (A – B)] cos (A + B)

= cos2 A + cos2 B – cos2 (A + B) – cos (A – B) cos (A + B)

= cos2 A + cos2 B – cos2 (A + B) – (cos2 A – sin2 B)

= (sin2 B + cos2 B) – cos2 (A + B)

= 1 – cos2 (A + B) = sin2 (A + B)

6. (d)

( ) ( )( ) ( )

2sin cos sin 2

2sin cos sin 2

− − −

− − −

= ( ) ( ) ( )( ) ( ) ( )

sin sin sin 2

sin sin sin 2

− + + − − − −

− + + − − − −

[ 2 sin A cos B = sin (A + B) + sin (A – B)]

= ( ) ( )( ) ( )

sin sin 2 sin 2

sin sin 2 sin 2

+ − − −

+ − − − =

sin

sin

7. (a)

cos 7x + cos 5x + cos 3x + cos x

= (cos 7x + cos 3x) + (cos 5x + cos x)

= 7x 3x 7x 3x

2cos cos2 2

+ −

5x x 5x x

2cos cos2 2

+ − +

= (2 cos 5x cos 2x) + (2 cos 3x cos 2x)

= 2 cos 2x [cos 5x + cos 3x]

= 2 cos 5x 3x 5x 3x

2x 2cos cos2 2

+ −

= 2 cos 2x [2 cos 4x. cos x] = 4 cos x cos 2x cos 4x

8. (a)

cosec A + sec A = cosec B + sec B

cosec A – cosec B = sec B – sec A

1 1 1 1

sin sin cosB cosAA B− = −

− −

=sinB sinA cosA cosB

sinAsinB cosAcosB

+ − + −

=

A B B A A B B A2cos sin 2sin sin

2 2 2 2

sinAsinB cosAcosB

+

=+

A Bcos

sin AsinB2

A B cos AcosBsin

2

cotA B

2

+ = tan A tan B

9. (b)

A, B & C are in A.P.

So, B = A C

2

+

sin A sinc

cosC cosA

−

− =

+ −

+ −

A C A C2cos sin

2 2

A C A C2sin sin

2 2

= cot +

A C

2= cot B

10. (c)

sin sin ( 2 )b a = +

sin

sin ( 2 )

a

b

=

+

sin sin ( 2 )

sin sin ( 2 )

a b

a b

+ + + =

− − +

2sin ( )cos

2cos ( )sin

+=

− + tan ( )cot = − +

cot

cot ( )

= −

+

11. (c)

cos 2x cos x

2 – cos 3x cos

9x

2

= 1 x 9x2cos2xcos 2cos cos3x

2 2 2

−

= 1 x x

cos 2x cos 2x2 2 2

+ + −

9x 9xcos 3x cos 3x

2 2

− + + −

= 1 5x 3x 5x 3xcos cos cos cos

2 2 2 2 2

+ − −



85

= 1 5x 15xcos cos

2 2 2

−

=

+ −

5x 15x 15x 5x1 2 2 2 22sin sin2 2 2

= 10x 5x

sin sin2 2

= 5x

sin5xsin2

12. (a)

sin2sin3sinsin =++

sin2sincos2sin2 =+

sin)1cos2(2sin =+ ….. (i)

Now cos2cos3coscos =++

cos2coscos2cos2 =+

cos)1cos2(2cos =+ ….. (ii)

Divide (i) by (ii)

tan2tan =

=2 2/ =

13. (a)

2cos cos3 cos5x x x− − ( )2cos cos3 cos5x x x= − +

2cos 2cos cos4x x x= − 2cos (1 cos4 )x x= −

xx 2sin2cos2 2=24cos sin 2x x= 2 316sin cosx x=

14. (a)

cos cosA m B= cos

1 cos

m A

B=

1 cos cos

1 cos cos

m A B

m A B

+ + =

− −=

2cos cos2 2

2sin sin2 2

A B B A

A B B A

+ −

+ −

−

+=

2cot

2cot

ABBA

Hence, 2

tan1

1

2cot

AB

m

mBA −

−

+=

+

15. (b)

sin A = sin B

sin A – sin B = 0

A B A B2sin cos

2 2

− +

= 0 ….. (i)

cos A = cos B

cos A – cos B = 0

A B A B2sin cos

2 2

− + −

= 0 ….. (ii)

From these 2 equations we can conclude that

A Bsin

2

−

= 0 ( it is in both the equations)

16. (c)

0coscoscos =++ yx coscoscos −=+ yx

cos2

cos2

cos2 −=

−

+ yxyx …..(i)

and sin sin sin 0x y + + = .sinsinsin −=+ yx

sin2

cos2

sin2 −=

−

+ yxyx ….. (ii)

Divide (i) by (ii)

−

+

−

+

2cos

2sin2

2cos

2cos2

yxyx

yxyx

sin

cos= cot

2cot =

+ yx

17. (b)

o

ooooo

70sin

70cos70sin470cos70cos470cot

+=+

cos70 2sin140

sin 70

o o

o

+=

cos70 2sin(180 40 )

sin70

o o o

o

+ −=

sin 20 sin 40 sin 40

sin70

o o o

o

+ +=

2sin30 cos10 sin 40

sin 70

o o o

o

+=

sin80 sin 40

sin70

o o

o

+=

2sin60 cos203

sin70

o o

o= =

18. (d)

sin8xcosx sin6xcos3x

cos2xcosx sin3xsin4x

−

−

=

12sin8xcosx 2sin6xcos3x

21

2cos2xcosx 2sin3xsin4x2

−

−

= ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

sin 8x x sin 8x x sin 6x 3x sin 6x 3x

cos 2x x cos 2x x cos 3x 4x cos 3x 4x

+ + − − + + −

+ + − − + + −

= sin9x sin7x sin9x sin3x

cos3x cosx cosx cos7x

+ − +

+ − +

[ cos (– x) = cos x]

= −

+

sin7x sin3x

cos3x cos7x=

+ −

+ −

7x 3x 7x 3x2cos sin

2 2

3x 7x 3x 7x2cos cos

2 2

= ( )

cos5xsin2x sin2x

cos2xcos5xcos 2x=

−

[ cos (– 2x) = cos 2x] = tan 2x

19. (d)

cos 6x + 6 cos 4x + 15 cos 2x + 10


86

= (cos 6x + cos 4x) + (5 cos 4x + 5 cos 2x) + (10 cos 2x +

10)

= (cos 6x + cos 4x) + 5 (cos 4x + cos 2x + 10 (cos 2x +

cos 0x)

=

10x 2x2cos cos

2 2+

6x 2x5 2cos cos

2 2+

2x 2x10 2cos .cos

2 2

= 2 cos 5x cos x + 5 (2 cos 3x cos x) + 10 (2 cos x cos x)

= 2 cos x [cos 5x + 5 cos 3x + 10 cos x]

Now,

cos6x 6cos4x 15cos2x 10

cos5x 5cos3x 10cosx

+ + +

+ +

= ( )2cosx cos5x 5cos3x 10cosx

cos5x 5cos3x 10cosx

+ +

+ + = 2 cos x

20. (d)

D

C

BA

BA=

+

+

coscos

sinsin

2 sin cos2 2

2cos cos2 2

A B A B

C

A B A B D

+ −

=+ −

tan2

A B C

D

+ =

Now, 2

2 tan2

sin ( )

1 tan2

A B

A BA B

+

+ =+

+

=2 2 2

2

22

( )1

C

CDD

C C D

D

=+

+

21. (c)

oooo 81tan63tan27tan9tan +−−

oooo 9cot27cot27tan9tan +−−=

)27cot27(tan)9cot9(tan oooo +−+=

= sin9 cos9 sin 27 cos27

cos9 sin9 cos27 sin 27

+ − +

cos(9 9 ) cos(27 27 )

sin9 cos9 sin 27 cos27

o o o o

o o o o

− −= −

2 2

sin18 sin54o o= − sin 54 sin 18

2sin 18 sin 54

o o

o o

−=

2cos36 sin18

2 4sin18 sin54

o o

o o= =

22. (a)

Given that tan 1 tan 2 = k

( )( )

1 2

1 2

cos

cos

−

+ = 1 2 1 2

1 2 1 2

cos cos sin sin

cos cos sin sin

+

−

= 1 2

1 2

1 tan tan 1 k

1 tan tan 1 k

+ +=

− −

23. (a)

+ +

+ +

cos3 2cos5 cos7

cos 2cos3 cos5 + sin 2 tan 3

= + +

+ +

cos3 cos7 2cos5

cos cos5 2cos3+ sin 2 tan 3

= 2cos5 .cos2 2cos5

2cos3 cos2 2cos3

+

+ + sin 2 tan 3

=

+

+

2cos5 cos2 1

2cos3 cos2 1 + sin 2 tan 3

= cos5 sin2 sin3

cos3 cos3

+

[cos5 = cos (2 + 3)]

= cos2 cos3 sin2 sin3 sin2 sin3

cos3

− +

= cos 2

24. (d)

sin sin2 sin4 sin5

cos cos2 cos4 cos5

A A A A

A A A A

+ + +

+ + +

= ( ) ( )( ) ( )

sin5 sin sin4 sin2

cos5 cos cos4 cos2

A A A A

A A A A

+ + +

+ + +

2sin3 cos2 2sin3 cos

2cos3 cos2 2cos3 cos

A A A A

A A A A

+=

+

( )( )

2sin3 cos2 cos

2cos3 cos2 cos

A A A

A A A

+=

+= tan3A

25. (c)

sin cos sin cos2 2 2 2

A B A B C D C D+ − + − +

=2

1

2[sinA + sinB+ sin C + sinD] =2

sin A + sin B + sinC + sin D = 4

A = B = C = D = 90°

So, sin1

sin sin sin2 2 2 2 4

A B C D=

26. (d)

2 sin (60° – x) cos (30° + x)

= sin (60° – x + 30° + x) + sin (60° – x – 30° – x)

= sin 90° + sin (30 – 2x)

= 1 + sin 30° cos 2x – cos 30° sin 2x

= 1 + cos2x 3 sin2x

2 2−



87

1. Trigonometric functions of 2A in terms of A.

(i) sin 2A = 2 sin A cos A

(ii) cos 2A = cos2 A – sin2 A

= 2 cos2 A – 1

= 1 – 2 sin2 A

(iii) tan 2A = 2

2tan A

1 tan A−

(iv) sin 2A = 2

2tan A

1 tan A+

(v) cos 2A = 2

2

1 tan A

1 tan A

−

+

(i) sin 2A = sin (A + A)

= sin A cos A + cos A sin A

= 2 sin A cos A

sin 2A = 2 sin A cos A

(ii) cos 2A = cos (A + A)

= cos A cos A – sin A sin A

= cos2 A – sin2 A

cos 2A = cos2 A – sin2 A

= cos2 A – (1 – cos2 A)

= 2 cos2 A – 1

cos 2A = cos2 A – sin2 A

= (1 – sin2 A) – sin2 A

= 1 – 2 sin2 A

(iii) tan 2A = tan (A + A)

= 2

tanA tanA 2tanA

1 tanAtanA 1 tan A

+=

− −

tan 2A = 2

2tan A

1 tan A−

(iv) sin 2A = 2sinAcosA

1

[ cos2 A + sin2 A = 1]

= +2 2

2sinAcosA

cos A sin A

= 2

2 2

2

2sin Acos A

cos Acos A sin A

cos A

+

= 2

2tan A

1 tan A+

(v) cos 2A = −

+

2 2

2 2

cos A sin A

cos A sin A

=

−

+

2 2

2

2 2

2

cos A sin A

cos Acos A sin A

cos A

= 2

2

1 tan A

1 tan A

−

+

------------------* * *------------------

(1) cos 2A = 1 – 2 sin2 A

2 sin2 A = 1 – cos 2A

sin2 A = 1– cos2

2

A

(2) cos 2A = 2 cos2 A – 1

2 cos2 A = 1 + cos 2A

cos2 A = 1 cos2

2

A+

2. Trigonometric functions of 3A in terms of A

(i) sin 3A = 3 sin A – 4 sin3 A

(ii) cos 3A = 4 cos3 A – 3 cos A

(iii) tan 3A = 3

2

3tanA tan A

1 3tan A

−

−

(i) sin 3A = sin (A + 2A)

= sin A cos 2A + cos A sin 2A

= sin A(1 – 2 sin2 A) + cos A 2 sin A cos A

[ cos 2A = 1 – 2 sin2 A and sin 2A = 2 sin A cos A]

= sin A – 2 sin3 A + 2 sin A (1 – sin2 A)

= 3 sin A – 4 sin3 A

Multiples and sub Multiples 8


88

sin 3A = 3 sin A – 4 sin3 A

(ii) cos 3A = cos (A + 2A)

= cos A cos 2A – sin A sin 2A

= cos A (2 cos2 A – 1) – sin A 2 sin A cos A

= 2 cos3 A – cos A – 2 cos A sin2 A

= 2 cos3 A – cos A – 2 cos A (1 – cos2 A)

= 4 cos3 A – 3 cos A

cos 3A = 4 cos3 A – 3 cos A

(iii) tan 3A = tan (A + 2A)

= tanA tan2A

1 tanAtan2A

+

−

= +

−

−−

2

2

2tan Atan A

1 tan A2tan A

1 tan A1 tan A

=− +

− −

3

2 2

tanA tan A 2tanA

1 tan A 2tan A

= 3

2

3tanA tan A

1 3tan A

−

−

tan 3A = 3

2

3tan A tan A

1 3tan A

−

−

3. Trigonometric functions of A in terms of A

2.

Replacing A by A

2 in the results of trigonometric

functions of 2A in terms of those of A, we get

(i) sin A = 2 sinA A

cos2 2

(ii) cos A = 2 2A A

cos sin2 2−

= 1 – 2 sin2A

2

= 2 cos2 A

2 – 1

(iii) 2 A 1 cosA

sin2 2

−=

(iv) 2 A 1 cosA

cos2 2

+=

(v) tan A = 2

A2tan

2A

1 tan2

−

(vi) sin A = 2

A2tan

2A

1 tan2

+

(vii) cos A =

2

2

A1 tan

2A

1 tan2

−

+

TRIGONOMETRIC FUNCTIONS OF

SOME PARTICULAR ANGLES

Values at 18°, 36°

Let x = 18°. Then 5x = 90°

2x + 3x = 90° 2x = 90° – 3x

sin 2x = sin (90° – 3x) = cos 3x

2 sin x cos x = 4 cos3 x – 3 cos x

2 sin x = 4 cos2 x – 3

2 sin x = 4 (1 – sin2 x) – 3

2 sin x = 1 – 4 sin2 x

4 sin2 x + 2 sin x – 1 = 0

Which is a quadratic in sin x

sin x = 2 4 16 2 20

8 8

− + − =

= 2 2 5 1 5

8 4

− − =

sin 18° = 1 5

4

−

Since 18° lies in the first quadrant, sin 18° is +ve. Hence

rejecting the –ve sign we get

sin18° = 5 1

4

−

Also cos2 x = 1 – sin2 x

cos2 18° = 1 – sin2 18° =

2

5 11

4

−−

= + −

−5 1 2 5

116

= ( )16 5 1 2 5 10 2 5

16 16

− + − +=

cos 18° = 10 2 5 10 2 5

16 4

+ + =

[Taking +ve sign before the square root, since 18° is in

1st quadrant and in 1st quadrant cosine is +ve]

(i) sin 72° = sin (90° – 18°)

= cos 18° = 10 2 5

4

+



89

(ii) cos 72° = cos (90° – 18°) = sin 18° = 5 1

4

−

(iii) cos 36° = 1 – 2 sin2 18° [ cos x = 1 – 2 sin2 x/2]

= 1 – 2

2

5 1

4

−

5 1

sin184

− =

= 1 – 2 + −

5 1 2 5

16 =

− + +=

8 6 2 5 5 1

8 4

(iv) sin 36° = − 21 cos 36 = +

−

2

5 11

4

= + +−

5 1 2 51

16=

−10 2 5

4 =

−10 2 5

4

(v) sin 54° = sin (90° – 36°) = cos 36° = 5 1

4

+

(vi) cos 54° = cos (90° – 36°) = sin 36° = 10 2 5

4

−

Values at 22.5° and 67.5°

tan 22.5° and cot 67.5°

tan2 = −

+

1 cos2

1 cos2

tan2 (22.5°)=1 cos45

1 cos45

−

+ =

12

12

−

+

= ( )22 1

2 12 1

−= −

+

tan 22.5° = 2 1−

cot 67.5° = cot (90° – 22.5°) = tan 22.5° = 2 1−

cot 22.5° and tan 67.5°

cot 22.5° = 1

tan22.5 =

12 1

2 1= +

−

tan (67.5°) = cot 22.5° = 2 1+


90

1. (sinA + cosA)2 =?

Sol. (sin A + cosA)2

= sin2A +cos2A + 2 sinA cosA = 1 + sin2A

2. tan A + cot A =?

Sol. tan A + cot A

= sin cos

cos sin

A A

A A+ =

2 2sin cos

sin cos

A A

A A

+

= 1

sin cosA A=

2

2sin cosA A =

2

sin2A = 2cosec2A

3. −

+

1 cos2

1 cos2

A

A =?

Sol. 1 cos2

1 cos2

A

A

−

+ =

( )− −

+ −

2

2

1 1 2sin

1 2cos 1

A

A

= 2

2

1 1 2sin

2cos

A

A

− +=

2

2

2sin

2cos

A

A= tan2A

4. cos4 A – sin4 A =?

Sol. cos4A – sin4A= (cos2A)2 – (sin2A)2

= (cos2A + sin2A) (cos2A – sin2A) = cos 2 A

5. −

=+

1 cos?

1 cos

A

A

Sol. 1 cos

1 cos

A

A

−

+=

2

2

1 1 2sin2

1 2cos 12

A

A

− −

+ −

= 2

2sin cos2 2

1 1 2sin2

A A

A− +

= 2

2sin cos2 2

2sin2

A A

A=

cos2

sin2

A

A = cot

2

A

6. sin

1 cos

A

A−=?

Sol. sin

1 cos

A

A−=

2

2sin cos2 2

1 1 2sin2

A A

A − −

= 2

2sin cos2 2

2sin2

A A

A = cot

2

A

7. 1sin2cos2 22 =− , then =

Sol. 1sin2cos2 22 =− 12cos2 =

o60cos

2

12cos == oo 30602 ==

8. +

sin2

1 cos2

A

A =?

Sol. sin2

1 cos2

A

A+=

2

2sin cos

1 2cos 1

A A

A+ −=

2

2sin cos

2cos

A A

A=

sin

cos

A

A= tan A

9 Find value of sin2A if cosA = 3

2

Sol. cos A = 3

2

now, sin A = 21 cos A− =

2

31

2

−

= 3

14

− = 1

2

Now sin2A = 2sinAcosA = 2×1

2×

3

2 =

3

2

10. If tan A = 5

12then cos 2A=?

Sol. tanA = 5

12

now, sin 2A = 2

2tan

1 tan

A

A+ sin 2A =

2

52

12

51

12

+

=

5

625

1144

+

=

5

6169

144

= 144 5

6 169

=

120

169

Examples



91

1. −

sin2

1 cos2

A

A =?

(a) cot A (b) tan A (c) cot 2A (d) tan 2A

2. cos4A – sin4 A =?

(a) sin 2A (b) 1 (c) cos 2A (d) 0

3. 2

2tan15

1 tan 15

− =?

(a) 3 (b) 1

3 (c) 1 (d) 0

4. (sinA - cosA)2 =?

(a) 1 + sin 2A (b) 1 – sin 2A

(c) sin A + cos A (d) tan 2A

5. sin2

1 cos2

A

A− =?

(a) tan A (b) tan 2A (c) cot A (d) cot 2A

6. cot A – tan A =?

(a) 2 tan 2A (b) 1 (c) sec 2A (d) 2 cot 2A

7. sin sin2

1 cos cos2

+

+ +=?

(a) tan θ (b) cot θ (c) 1 (d) tan 2θ

8. If cos A = 5

13 then sin 2A =?

(a) 12

13 (b)

24

13 (c)

120

169 (d)

13

24

9. If tan A = 1

3, then tan2A =?

(a) 2

3 (b) 3 (c) 2 3 (d) 1

10. If cos 2A = 13

36 then cos A =?

(a) 13

18 (b)

7

6 2 (c)

7

6 (d) 1

11. If sinA = 3

5then tan 2A =?

(a) 4

5 (b)

7

24 (c)

5

4 (d)

24

7

12. sin

1 cos

A

A+=?

(a) tan2

A (b) tan A (c) cot

2

A (d) cot A

13. 1

cot tan2 2 2

A A −

=?

(a) sec 2A (b) tan A (c) cosec 2A (d) cot A

14. tan cot2 2

A A+ =?

(a) sec 2A (b) tan A (c) cosec 2A (d) cot A

15. If ,2

tan t=

then 2

2

1

1

t

t

+

−is equal to

(a) cos (b) sin (c) sec (d) 2cos

16. If ,2cossin =+ AA then =A2cos

(a) 4

1 (b)

2

1 (c)

2

1 (d)

2

3

17. =− AAAA cossin2cossin2 33

(a) A4sin (b) A4sin2

1

(c) A4sin4

1 (d) None of these

18. If tan A = 1 then (sinA + cosA)2=?

(a) 1 (b) 2 (c) 1

4 (d)

1

2

19. If cos A = 1

2 then tan2A =?

(a) 3 (b) 1

3 (c) 3− (d)

1

3−

20. +

−

sin cos

sin cos

A A

A A =?

(a) 1 sin2

1 sin2

A

A

+

− (b)

1 sin2

1 sin2

A

A

−

+

Exercise - 1


92

(c) sin 2A (d) 1

sin2A

21. =15cos

(a) 2

30cos1 + (b)

2

30cos1 −

(c) 2

30cos1 + (d)

2

30cos1 −

22. 2

tanA

is equal to

(a) A

A

sin1

sin1

+

− (b)

A

A

sin1

sin1

−

+

(c)A

A

cos1

cos1

+

− (d)

A

A

cos1

cos1

−

+

23. If 2cos2A = cos245°+ sin60° then find the value of cos2A

(a) 3 1

2

+ (b)

3 1

2

− (c) - 1 (d) 1

24. −

+

cot tan

cot tan

A A

A A=?

(a) sec 2A (b) sin 2A (c) cos 2A (d) cosec 2A

25.

2 2

2 2

1 1sin 22 cos 22

2 21

sin 22 cos 222 2

−

+

=?

(a) 1

2 (b) 2 (c) 1 (d)

1

2−



93

1. 1 cos2 sin2

1 cos2 sin2

A A

A A

− +

+ + =?

(a) tan A (b) cot A (c) tan 2A (d) cot 2A

2. cosec 2A + cot 2A =?

(a) cot 2A (b) tan 2A (c) tan A (d) cot A

3. sin2 cos

1 cos2 1 cos

A A

A A

+ +=?

(a) tan 2A (b) cot 2A (c) tan2

A (d) cot

2

A

4. 2sin sin2

2sin sin2

+

− =?

(a) cot2

2

(b) tan2

2

(c) cot

2

(d) tan

2

5. If ,1

2

1cos

+=

aa then the value of 3cos is

(a)

+

3

3 1

8

1

aa (b)

+

aa

1

2

3

(c)

+

3

3 1

2

1

aa (d)

+

3

3 1

3

1

aa

6. Which of the following is rational

(a) 15sin (b) 15cos

(c) 15cos15sin (d) 75cos15sin

7. =++

−1

cossin

3cos3sin

(a) 2sin2 (b) 2cos2 (c) 2tan (d) 2cot

8. If ,sin

cos1tan

B

BA

−= find A2tan in terms of Btan and

show that

(a) BA tan2tan = (b) BA 2tan2tan =

(c) BBA tan2tan2tan 2 += (d) None of the above

9.

+ +

+ −

1 sin2 cos2

1 sin2 cos2=?

(a) tan (b) cot2 (c) cot (d) tan2

10. If cosA = 3

2, then cos3A =?

(a) 0 (b) 1 (c) 1

2 (d)

1

2−

11. If ,5

3sin

−= where ,

2

3 then cos

2

=

(a) 10

1− (b)

10

1 (c)

10

3 (d)

10

3−

12. If cos45° = 1

2then sin22

1

2 =?

(a) 2 2

2

+ (b)

−2 2

2

(c) 3 1

2 2

− (d) None of these

13. If ,5

1cossin =+ xx then x2tan is

(a) 17

25 (b)

25

7 (c)

7

25 (d)

7

24

14. sin2 (A + B)- sin2(A -B) =?

(a) 1 (b) sin 2A

(c) sin 2A sin 2B (d) sin A sin B

15. tan A cot2

A-1 =?

(a) sec A (b) cos A (c) cosec A (d) cos2

A

16. 3 3sin 15 cos 15

sin15 cos15

−

− =?

(a) 3

5 (b)

4

5 (c) 1 (d)

5

4

17. If cos A = 3

5 and cosB =

4

5 then find the value of cos

2

A B−

where both A and B are positive acute angles.

Exercise - 2


94

(a) 25

49 (b)

24

49 (c)

49

50 (d)

49

25

18. Let .4

0

x then =− xx 2tan2sec

(a)

−

4tan

x (b)

− x

4tan

(c)

+

4tan

x (d)

+

4tan2

x

19. 1 + tan 2Atan A =?

(a) sec2A (b) sin2A (c) cot2A (d) cosec2A

20. =− A

A

sin1

cos

(a) AA tansec − (b) AA cotcosec +

(c)

−

24tan

A (d)

+

24tan

A

21. −

cos2

1 sin2

A

A=?

(a) tan (450 + A) (b) tan (450 - A)

(c) sec A – tan A (d) 1

22. sec2 A (1 + sec2A) =?

(a) cosec 2A (b) sec 2A

(c) 2 sec 2A (d) cosec A – cot A

23. 2 cos (45°+ x) cos (45°- x) =?

(a) sin x (b) cos 2x (c) sec x – cos x (d) 0

24.

tan tan4 4

tan tan4 4

+ − −

+ + −

=?

(a) tan 2α (b) sin 2α (c) cot α (d) cot 2α

25. 2 1 2 2cos8+ + + =?

(a) 2 cos (b) 2 sin (c) 2 cos2 (d) 1



95

Answer key

1 a 2 c 3 b 4 b 5 c

6 d 7 a 8 c 9 b 10 b

11 d 12 a 13 d 14 c 15 a

16 b 17 b 18 b 19 c 20 a

21 a 22 c 23 b 24 c 25 d

1. (a)

sin2

1 cos2

A

A−=

( )2

2sin cos

1 1 2sin

A A

A− −

= 2

2sin cos

1 1 2sin

A A

A− +=

2

2sin cos

2sin

A A

A=

cos

sin

A

A = cot A

2. (c)

cos4A –sin4A

= (cos2A + sin2A) (cos2A – sin2A) = 1× cos 2A = cos2A

3. (b)

2

2tan15

1 tan 15

− = tan (2×15°) = tan30°=

1

3

4. (b)

(sin A - cosA)2 = sin2A + cos2A –2 sinA cosA

= 1 – sin2A

5. (c)

( )2

sin2 2sin cos

1 cos2 1 1 2sin

A A A

A A=

− − −

= 2

2sin cos

1 1 2sin

A A

A− + =

2

2sin cos

2sin

A A

A=

cos

sin

A

A = cotA

6. (d)

cotA – tanA = cos

sin

A

A -

sin

cos

A

A

= 2 2cos sin

sin cos

A A

A A

−=

cos2

sin cos

A

A A =

2cos2

2sin cos

A

A A

= 2cos2

sin2

A

A = 2cot 2A

7. (a)

sin sin2

1 cos cos2

+

+ + =

2

sin 2sin cos

1 cos 2cos 1

+

+ + −

= ( )( )

sin 1 2cos

cos 1 2cos

+

+ =

sin

cos

= tan θ

8. (c)

cosA = 5

13

sinA =

25

113

−

= 25

1169

− = 144

169=

12

13

sin 2A = 2sinAcosA = 2×12

13×

5

13=

120

169

9. (b)

tanA = 1

3

tan2A = 2

2tan

1 tan

A

A−=

2

1 21

3 311 11 3

3

= −−

=

1

32

3

= 2 3

2 3

= 3

10. (b)

cos 2A = 13

36 [ 2cos2A- 1 = cos2A]

2cos2A – 1 = 13

36 2cos2A = +

131

36=

49

36

cos2A = 49

2 36 cos A =

49

2 36=

7

6 2

11. (d)

sinA = 3

5,

p

h

Solutions

Exercise - 1


96

base = 2 25 3− = 25 9− = 16 = 4

tan A = p

b=

3

4

tan 2A = 2

2tan

1 tan

A

A− =

2

32

4

31

4

−

=

3

29

116

−

=

3

27

16

= 3 16

7 2

=

24

7 =

33

7

12. (a)

sin

1 cos

A

A+

= 2

2sin cos2 2

1 2cos 12

A A

A+ −

= 2

2sin cos2 2

2cos2

A A

A=

sin2

cos2

A

A= tan

2

A

13. (d)

1

2cot tan

2 2

A A −

=

1

2

cos sin2 2

sin cos2 2

A A

A A

−

=

2 2cos sin2 2

2sin cos2 2

A A

A A

− =

cos

sin

A

A

2 2cos cos sin2 2

sin 2sin cos2 2

A AA

A AA

= −

=

= cot A

14. (c)

tan2

A+ cot

2

A =

sin2

cos2

A

A +

cos2

sin2

A

A

=

2 2sin cos2 2

cos sin2 2

A A

A A

+= 1

cos sin2 2

A A

= 2

2cos sin2 2

A A =

2

sin A= 2cosecA

15. (a)

tan2

t= ⇒

2

2

1

1

t

t

+

− =

2

2

1 tan2 cos

1 tan2

−

=

+

16. (b)

2cossin =+ AA

On squaring both the sides

oAA 90sin12sin22sin1 ===+

oA 902 = or oA 45=

Now, 2

1

2

1)45(coscos

2

22 =

== oA

17. (b)

AAAA cossin2cossin2 33 − )sin(coscossin2 22 AAAA −=

2sin cos cos2A A A= = sin2 cos2A A = 1

sin 42

A

18. (b)

tanA = 1

(sin A + cos A)2

= sin2 A + cos2A + 2sinA cosA

= 1 + sin 2A = 1 + 2

2tan

1 tan

A

A+ = 1+

2

2 1

1 1

+=2

Or

tan A = 1 A = 450

Now (sin450 + cos450)2 =

21 1

2 2

+

= ( )2

2 =2

19. (c)

cos A = 1

2, A = 600

tan A = 3

tan2A = 2

2tan

1 tan

A

A−=

( )2

2 3

1 3

−

= 2 3

1 3−=

2 3

2−= − 3

20. (a)

sin cos

sin cos

A A

A A

+

−= ( )

( )

2

2

sin cos

sin cos

A A

A A

+

−

=

2 2

2 2

sin cos 2sin cos

sin cos 2sin cos

A A A A

A A A A

+ +

+ −=

1 sin2

1 sin2

A

A

+

−

21. (a)

1 cos(2 15 )

cos152

oo + = =

1 cos30

2

o+

( )015cos o

22. (c)

sin( / 2)

tan2 cos( / 2)

A A

A

=

=



97

(1 cos ) / 2 1 cos

(1 cos ) / 2 1 cos

A A

A A

− − =

+ +

23. (b)

2cos2 A = cos2 45°+sin60°

=

21

2

+ 3

2=

1

2+

3

2=

1 3

2

+

cos2A = 2cos2 A – 1

= 1 3

2

+ - 1 =

1 3 2

2

+ −=

3 1

2

−

24. (c)

−

+

cot tan

cot tan

A A

A A=

−

+

1tan

tan1

tantan

AA

AA

= −

+

2

2

1 tan

1 tan

A

A= cos 2A

25. (d)

2 2

2 2

1 1sin 22 cos 22

2 21

sin 22 cos 222 2

−

+

+ = 2 2sin cos 1

= − 2 21 1sin 22 cos 22

2 2 =

− −

2 21 1cos 22 in 22

2 2s

= - cos450 =1

2−

Answer key

1 a 2 d 3 c 4 a 5 c

6 c 7 a 8 a 9 c 10 a

11 a 12 b 13 d 14 c 15 a

16 d 17 c 18 b 19 a 20 d

21 a 22 c 23 b 24 b 25 a

1. (a)

1 cos2 sin2

1 cos2 sin2

A A

A A

− +

+ +=

( )2

2

1 1 2sin 2sin cos

1 2cos 1 2sin cos

A A A

A A A

− − +

+ − +

= 2

2

1 1 2sin 2sin cos

2cos 2sin cos

A A A

A A A

− + +

+=

2

2

2sin 2sin cos

2cos 2sin cos

A A A

A A A

+

+

= ( )( )

2sin sin cos

2cos cos sin

A A A

A A A

+

+=

sin

cos

A

A= tan A

2. (d)

cosec 2A + cot 2A

= 1 cos2 1 cos2

sin2 sin2 sin2

A A

A A A

++ =

= 21 2cos 1

2sin cos

A

A A

+ −=

22cos

2sin cos

A

A A =

cos

sin

A

A= cot A

3. (c)

sin2 cos

1 cos2 1 cos

A A

A A

+ +=

( )

++ −2

2sin cos cos

1 cos1 2cos 1

A A A

AA

= ( )2

2sin cos cos sin

1 cos2cos 1 cos

A A A A

AA A

=

+ +

= 2

2sin cos2 2

1 2cos 12

A A

A+ −

= 2

2sin cos2 2

2cos2

A A

A= tan

2

A

4. (a)

2sin sin2

2sin sin2

+

− =

2sin 2sin cos

2sin 2sin cos

+

−

= ( )( )

2sin 1 cos

2sin 1 cos

+

−

= 1 cos

1 cos

+

− =

2

2

1 2cos 12

1 1 2sin2

+ −

− −

Exercise - 1


98

=

− +

2

2

2cos2

1 1 2sin2

=

2

2

2cos2

2sin2

= cot2

2

5. (c)

cos3 = 34cos 3cos −

= 3

3

1 1 1 14 3

22a a

a a

+ − +

=

21 1 1

32

a aa a

+ + −

= 3

3

1 1

2a

a

+

6. (c)

=−

=−=22

13)3045sin(15sin ooo irrational

22

13)3045cos(15cos

+=−= ooo =irrational

)15cos15sin2(2

115cos15sin oooo =

1 1 1 1

sin302 2 2 4

o= = = = rational

ooooo 15sin15sin15sin75cos15sin 2==

8

324

22

132

−=

−= = irrational

7. (a)

sin3 cos3

sin cos

−

+ + 1

= 3 33sin 4sin (4cos 3cos )

sin cos

− − −

+ + 1

= 3 33(sin cos ) 4(sin cos )

sin cos

+ − +

+ + 1

=2 2(sin cos ){3 4(sin sin cos cos )}

sin cos

+ − − +

++1

1)cossin1(43 +−−= 2sin2cossin4 ==

8. (a)

1 cos

tansin

BA

B

−= =

22sin ( / 2)

2sin( / 2) cos( / 2)

B

B B

= tan2

B A =

2

B, 2A = B BA tan2tan =

9. (c)

+ +

+ −

1 sin2 cos2

1 sin2 cos2 =

( )( )

+ +

− +

1 cos2 sin2

1 cos2 sin2

2

2

2cos 2sin cos

2sin 2sin cos

+=

+

( )( )

+=

+

2cos cos sin

2sin cos sin

=

=

coscot

sin

10. (a)

cos A = 3

2

sin A = =

2

31

2

−

= 3

14

− = 1

4 =

1

2

cos 3A = 4cos3A – 3cosA

= 4

3

3

2 - 3×

3

2

= 4 ×3 3 3 3 3 3 3 3

8 2 2 2− = − = 0

0r cos A = 3

2 A = 300

Now cos3A = cos 900 = 0

11. (a)

2

cos1)2/cos(

+−=

cos = 21 sin − − [ lies in IIIrd Quadrant]

5

4

25

91 −=−−=

10

1

2

5

41

)2/cos( −=

−

−=

12. (b)

cos 45° = 1

2

1- 2sin245

2

=

1

2 [ 1 – 2 sin2

2

A= cosA]

2sin2 45

2

=

11

2− 2sin2

45

2

=

2 1

2

−

2sin245

2

=

( )2 2 1

2 2

−

=

2 2

2

−

sin 221

2 =

2 2

4

− =

−2 2

2

13. (d)

5

1cossin =+ xx



99

25

1cossin2cossin 22 =++ xxxx

25

242sin

−=x and

25

72cos

−=x

7

242tan =x

14. (c)

sin2(A + B) – sin2 (A - B)

= {sin (A + B) + sin (A - B)} {(sin (A + B) – sin(A - B)}

= (sin A cos B + cos A sin B + sin A cos B – cosA sin B) + {

sin A cos B + cos A sin B – (sin A cos B – cos A sin B)}

= (2sinAcosB) (2cosAsinB)

= (2 sin A cos A) (2 cos B sin B)

= sin 2A sin 2B

15. (a)

tan A cot2

A - 1=

2

2tan2 cot 1

21 tan

2

AA

A−

−

= −

− 2

2tan cot2 2 1

1 tan2

A A

A= −

− 2

21

1 tan2

A

=

2

2

2 1 tan2

1 tan2

A

A

− +

−

=

2

2

1 tan2

1 tan2

A

A

+

−

= 1

cos A= sec A

16. (d)

3 3sin 15 cos 15

sin15 cos15

−

−

=( )( )

( )

2 2sin15 cos15 sin 15 cos 15 sin15 cos15

sin15 cos15

− + +

−

= 1 + sin15° cos15° = 1 + 2sin15 cos15

2

= 1+ sin30

2

=

11

4+ =

5

4

17. (c)

cos A = 3

5

sin A = 21 cos A− =

23

15

−

= 4

5

and cos B = 4

5

sin B = 21 cos B− =

24

15

−

= 3

5

cos (A - B) = 2cos2−

2

A B -1

Now,

cos A cos B + sin A sin B= 2cos2−

2

A B – 1

3 4 4 3

5 5 5 5 + = 2cos2

− 2

A B – 1

2cos2 −

2

A B – 1 =

24

25

cos2 −

2

A B =

49

50 cos

− 2

A B =

7

5 2

18. (b)

sec2 tan2x x− = 1 sin 2

cos2

x

x

−

= 2

2 2

(cos sin )

(cos sin )

x x

x x

−

− =

cos sin 1 tan

cos sin 1 tan

x x x

x x x

− −=

+ +

= tan tan

4

1 tan tan4

x

x

−

+

= tan4

x −

19. (a)

1+ tan2A tanA = 1 + 2

2tan

1 tan

A

A− tanA

= 1+ 2

2

2tan

1 tan

A

A− =

2 2

2

1 tan 2tan

1 tan

A

A

− +

−

= 2

2

1 tan

1 tan

A

A

+

−=

2

2

2

2

sin1

cossin

1cos

A

AA

A

+

−

= 2 2

2 2

cos sin

cos sin

A A

A A

+

−=

1

cos2A= sec2A

20. (d)

cos

1 sin

A

A−

= 2

cos (1 sin )

cos

A A

A

+ =

(1 sin )

cos

A

A

+

2

cos sin2 2

cos sin cos sin2 2 2 2

A A

A A A A

+

=

+ −


100

= cos sin

2 2

cos sin2 2

A A

A A

+

−2

tan1

2tan1

A

A

−

+

=

+=

24tan

A

21. (a)

cos2

1 sin2

A

A− =

2 2

2 2

cos sin

cos sin 2sin cos

A A

A A A A

−

+ −

= ( )( )

( )2

cos sin cos sin

cos sin

A A A A

A A

+ −

−

= cos sin

cos sin

A A

A A

+

−=

+

−

1 tan

1 tan

A

A= tan (450 + A)

22. (c)

sec2 A (1+sec2A) = 2

1

cos A

+

11

cos2A

= 2

1

cos A

+

cos2 1

cos2

A

A=

2

1

cos A

− +

22cos 1 1

cos2

A

A

= 2

2

1 2cos

cos2cos

A

AA=

2

cos2A = 2sec2A

23. (b)

2cos (45° + x) cos (45° - x)

= 2(cos45°cosx – sin450sinx) (cos45°cosx + sin45°sinx)

= 2 1 1

cos sin2 2

x x

−

1 1cos sin

2 2x x

+

= 2×1 1

2 2 (cos x – sin x) (cos x + sin x)

= 2

2 (cos2x – sin2x) = cos2x

24. (b)

tan tan4 4

tan tan4 4

+ − −

+ + −

=

+ −−

− ++ −

+− +

1 tan 1 tan

1 tan 1 tan1 tan 1 tan

1 tan 1 tan

=

( ) ( )( )( )

( ) ( )( )( )

+ − −

+ −

+ + −

+ −

2 2

2 2

1 tan 1 tan

1 tan 1 tan

1 tan 1 tan

1 tan 1 tan

= ( ) ( )

( ) ( )

+ − −

+ + −

2 2

2 2

1 tan 1 tan

1 tan 1 tan

=

+ 2

2tan

1 tan= sin 2α

25. (a)

2 2 2(1 cos8 )+ + +

= 22 2 2(2cos 4 )+ + 2 8

1 cos8 2cos2

+ =

= ( )22 2 4cos 4+ + = ( )2 2 1 cos4+ + +

= ( )22 2 2cos 2+ = 2 2cos2+ = ( )2 1 cos2+

= ( )22 2cos = 2 cos



101

Some Important Results

1. cos A cos (60° – A) cos (60° + A = 1

4cos 3A

2. sin A sin (60° – A) sin (60° + A) = 1

4sin 3A

3. tan tan (60° – ) tan (60° + ) = tan 3

4. sin2 + sin2(600 - ) + sin2(600 + ) = 3

2

5. cos2 + cos2(600 - ) + cos2(600 + ) = 3

2

6. cos cos 2 cos 4 ……… cos2n-1 = n

n

sin2

2 sin

7. sin + sin (+) + sin (+) +.... + sin (+n−) =

sin2 sin ( 1)

2sin

2

+ −

n

n

1. cos A cos (60° – A) cos (60° + A = 1

4cos 3A

Proof:

cos A cos (60° – A) cos (60° + A)

= cos A (cos2 60° – sin2 A)

[ cos (A + B) cos (A – B) = cos2 A – sin2 B]

= ( )2 21 1cosA sin A cosA 1 cos A

4 4

− = − −

= 23cosA cos A

4

− +

= 1

4cos A (–3 + 4 cos2 A)

= 1

4 (4 cos3 A – 3 cos A)

= 1

4cos 3A

2. sin A sin (60° – A) sin (60° + A) = 1

4sin A

Proof:

sin A sin (60° – A) sin (60° + A)

= sin A (sin2 60° – sin2 A)

[ sin (A + B) sin (A – B) = sin2 A – sin2 B]

= ( )2 23 1sinA sin A sinA 3 4sin A

4 4

− = −

= 1

4 (3 sin A – 4 sin3 A)

= 1

4sin 3A

3. tan tan (60° – ) tan (60° + ) = tan 3

4. sin2 + sin2(600 - ) + sin2(600 + ) = 3

2

Proof:

sin2 + sin2 (60° – ) + sin2 (60° + )

= sin2 + (sin 60° cos – cos 60° sin )2 +

(sin 60° cos + cos 60° sin )2

= sin2 + 2 2

3 1 3 1cos sin cos sin

2 2 2 2

− + +

= sin2 + 2 2 2 23 1 3 1

cos sin cos sin4 4 4 4

+ + +

= 2 23 3 3

sin cos2 2 2

+ =

5 cos2 + cos2 (60° – ) + sin2 (60° + ) =3

2

Proof:

cos2 + cos2 (60° – ) + sin2 (60° + )

= cos2 + (cos 60° cos + sin 60° sin )2 +

(cos 60° cos – sin 60° sin )2

= cos2 +

2 2

cos 3 cos 3sin sin

2 2 2 2

+ + −

= cos2 + 2 2

2cos 3 3 cossin sin cos

4 4 2 4

+ + +

23 3sin sin cos

4 2+ −

= 2 23 3 3

cos sin2 2 2

+ =

6. cos cos 2 cos 4 ……… cos2n-1 = n

n

sin2

2 sin

Proof:

Special Properties 9


102

cos cos 2 cos 4 ……… cos2n-1

= 1

22sin

sin cos cos 2 cos 4 …… cos2n-1

= 2

1

2 sin2 sin 2 cos 2 cos 4 ……… cos2n-1

= 3

1

2 sin2 sin 4 cos 4 ……… cos2n-1

Similarly

= n 1

1

2 sin− sin2n-1 cos2n-1

= n

1

2 sin2sin2n-1 cos2n-1

= n

1

2 sinsin (2×2n-1) =

n

n

sin2

2 sin

Some Standard Identities in Triangle

1. tan A + tan B + tan C = tan A tan B tan C

2. tan2

Atan

2

B+ tan

2

Ctan

2

B+ tan

2

Ctan 1

2

A=

3. sin 2A +sin 2B + sin 2C = 4 sin A sin B sin C

4. cos 2A + cos 2B + cos 2C = -1 – 4cos A cos Bcos C

5. cos A + cos B + cos C = 1 + 4 sin 2

Asin

2

Bsin

2

C

6. sin A + sin B + sin C = 4 cos 2

A cos

2

Bcos

2

C

1. tan A + tan B + tan C = tan A tan B tan C

Proof: in Δ ABC, we have A + B + C =

A + B = - C tan (A+B) = tan (-c)

tan tan

1 tan tan

A B

A B

+=

−- tan C

tan A + tan B = - tan C + tan A tan B tan C

tan A + tan B + tan C = tan A tan B tan C

2. tan2

Atan

2

B+ tan

2

Ctan

2

B+ tan

2

Ctan 1

2

A=

Proof:

Since A + B + C = , we have 2 2 2 2

A B C+ = −

tan tan cot

2 2 2 2 2

A B C C + = − =

tan tan

12 2

1 tan tan tan2 2 2

A B

A B c

+

=

−

tan2

Atan

2

C+ tan

2

Btan

2

C = 1 - tan

2

Atan

2

B

3. sin 2A +sin 2B + sin 2C = 4 sin A sin B sin C

Proof : (sin2A+sin 2B) + sin 2C = 2 sin (A+B) cos (A-B) + sin 2C

= 2 sin (-C) cos (A-B) + sin 2C

= 2 sin C cos (A-B) + 2 sin C cos C

= 2 sin C [cos (A-B) + cosC]

= 2 sin C [cos (A-B) + cos {-(A+B)]

= 2sin C[cos (A-B) – cos (A+B)]

= 2 sin C ×2 sin A sin B = 4 sin A sin B sinC

4. cos 2A + cos 2B + cos 2C = -1 – 4cos A cos Bcos C

Proof: (cos 2A + cos 2B) + cos 2C

= 2 cos (A+B) cos (A-B) + 2cos2 C-1

=2 cos (-C) cos (A-B) + 2cos2 C – 1

= -2 cos C cos (A-B) + 2cos2 C-1

=- 2 cos C [cos (A-B) –cos C]-1

=- 2 cos C [cos (A-B) –cos {-(A+B)} ]-1

= - 2 cos C [cos (A-B) + cos (A+B)] – 1

= - 1-4 cos A cos B cos C

5. cos A + cos B + cos C = 1 + 4 sin 2

Asin

2

Bsin

2

C

Proof:

(Cos A + cos B) + cos C -1

= 2cos cos cos2 2

A B A B+ −+ C – 1

= 2cos 2 2

C −

cos 2

A B−+ cos C – 1

= 2sin 2

C cos

2

A B−+1 – 2sin2

2

C - 1

= 2sin 2

Ccos

22sin2 2

A B C−−

= 2sin 2

Ccos sin

2 2

A B C− −

= 2sin 2

Ccos sin

2 2 2

A B A B − + −

= 2sin 2

Ccos cos

2 2

A B A B− + −

= 2sin 2

C2sin sin 4

2 2

A B =

sin

2

Acos

2

Bsin

2

C

6. sin A + sin B + sin C = 4 cos 2

A cos

2

Bcos

2

C

Proof :



103

(sin A + sin B) + sin C

= 2sin cos2 2

A B A B+ −+ sin C

= 2sin cos sin2 2 2

C A B − − +

C

= 2cos 2

Ccos

2

A B− + 2sin cos

2 2

C C

= 2cos 2

Ccos sin

2 2

A B C− +

= 2cos cos sin2 2 2 2

C A B A B − + + −

= 2cos cos cos2 2 2

C A B A B− + +

= 4cos cos cos2 2 2

A B C


104

1. cos 50° cos 70° cos 10° =?

Sol. cos 10°cos 50° cos 70°

= cos 100 cos( 600 – 100) cos(600 + 100)

= 1

4 cos 30°=

1 3 3

4 2 8 =

2. sin5 sin65 sin55

cos75

=?

Sol. sin5 sin65 sin55

cos75

=

( ) ( )sin5 sin 60 5 sin 60 5

cos75

+ −

( )sin 3 51

4 cos75

=

=

1 sin15

4 cos75

=

1 sin15 1

4 sin15 4

=

3. tan 12° cot 18° tan 48° cot 54°=?

Sol. tan 12° tan 48° cot 18° cot 54°

= tan12° tan (60°-12°) tan (60°+12°) cot 54°

= tan (3 × 120) cot 54°

= tan 36° cot 54° = 1

4. sin2 10° + sin2 70° + sin2 50° =?

Sol. sin2 10° + sin2 70° + sin2 50°

= sin2 10° + sin2 (60° + 10°) + sin2 (60° 10°) = 3

2

5. cos2 23° + cos2 83° + cos2 37° + sin2 60°

Sol. cos2 23° + cos2 83° + cos2 37° + sin2 60°

= cos223°+ cos2 (60°+23°) + cos2 (60° - 23°) + sin2 60°

=

2

3 3

2 2

+

= 3 3 9

2 4 4+ =

6. cos 20° cos 40° cos 60° cos 80°

Sol. (cos 20° cos 40° cos 80°) cos 60°

1st method:

= 3 4

sin160 1 1 1

2 162 sin20 2

= =

2nd method:

= cos3 20 1

4 2

=

cos60 1 1

4 2 16

=

7. 2 3

cos cos cos7 7 7

Sol. 2 3cos cos cos

7 7 7

= 2

42sin

37 cos7

2 2 sin7

=

4 3 4 3sin sin

7 7 7 7

8sin7

+ + −

=

sin sin178

8sin7

+

=

8. Find the value of cos 2 4 6

cos cos7 7 7

+ +

Sol. S = cos 2 4 6

cos cos7 7 7

+ +

=

sin 337

cos7 7

sin7

+

=

3 4 72sin cos sin sin

17 7 7 7

2 2 22sin 2sin

7 7

−

= = −

9. sin + sin 3 + sin 5 +..... + sin(2n-1) =?

Sol. sin + sin3 + sin5 +....+sin(2n-1)

=

2sin

2 (2 1)sin

2 2sin

2

nn

+ −

= 2sin

sin

n

10. If A + B + C = 180°,then cos2A +cos2B +cos2C=

Sol. 2 2 2cos cos cosA B C+ +

= 1 cos2 1 cos2 1 cos2

2 2 2

A B C+ + ++ +

= ( )1 3

cos2 cos2 cos22 2

A B C= + + +

( )1 3

1 4cos cos cos2 2

A B C= − − +

= 1 – 2cosA cos B cos C

11. In triangle ABC, cos2A+ cos2B-cos2C =?

Sol. 2 2 2cos cos cosA B C+ −

Examples



105

= 2 2 2cos sin sinA C B+ −

= ( ) ( )2cos sin inA C B s C B+ + −

= 1-sin2A + sin A sin (C -B)

= 1-sin A [sinA – sin(C-B)]

= 1-sin A [sin(B+C)-sin(C-B)] = 1-2 sin A sin B cosC

12. In triangle ABC,

sin (B+C-A) +sin (C + A - B) + sin (A + B -C) =?

Sol. sin (B + C – A ) + sin (C + A - B) + sin (A + B - C)

= sin ( 2A − )+ sin( 2B − ) + sin ( 2C − )

= sin 2A + sin 2B + sin 2C

= 4 sin A sin B sin C


106

1. tan 20° tan 80° cot 50° = ?

(a) 3 (b) 1

3 (c) 2 3 (d)

1

2 3

2. Find the value of 1 – sin 10° sin 50° sin 70°

(a) 5

8 (b)

7

8 (c)

1

4 (d)

3

4

3. Find the value of cos 20° cos 40° cos 60° cos 80°

(a) 1

16 (b)

1

8 (c)

3

16 (d)

1

4

4. 1 1

cos15 cos7 cos822 2

=?

(a) 1

8 (b)

1

4 (c)

3

8 (d)

1

2

5. tan 6° tan 42° tan 66° tan 78° = ?

(a) 1 (b) 2 (c) 3 (d) 4

6. Find the value of

sin 20° sin 40° sin 60° sin 80°

(a) 1

16 (b)

2

16 (c)

3

16 (d)

1

4

7. sin 12° sin 48° sin 54° = ?

(a) 1

4 (b)

3

4 (c)

3

8 (d)

1

8

8. cot 15° cot 25° cot 35° cot 85° = ?

(a) 1 (b) 0 (c) 2 (d) –1

9. Find the value of cos 10° cos 30° cos 50° cos 70°

(a) 1

16 (b)

2

16 (c)

3

16 (d)

1

4

10. Find the value of sin 6° sin 42° sin66° sin 78°

(a) 1

16 (b)

1

8 (c)

3

16 (d)

1

4

11. If ,180ozyx =++ then zyx 2cos2cos2cos −+ is equal to

(a) 4sin sin sinx y z (b) 1 4sin sin cosx y z−

(c) 4 sin x sin y sin z - 1 (d) cos A cos B cos C

12. If ,2 =++ then

(a) 2

tan2

tan2

tan2

tan2

tan2

tan

=++

(b) 12

tan2

tan2

tan2

tan2

tan2

tan =++

(c) 2

tan2

tan2

tan2

tan2

tan2

tan

−=++

(d) None of these

13. A, B, C are the angles of a triangle, then

=−++ CBACBA coscoscos2sinsinsin 222

(a) 1 (b) 2 (c) 3 (d) 4

14. If A, B, C are angles of a triangle, then

CBA 2sin2sin2sin −+ is equal to

(a) CBA coscossin4 (b) Acos4

(c) AA cossin4 (d) CBA sincoscos4

15. If ,180oCBA =++ then the value of

2cot

2cot

2cot

CBA++ will be

(a) 2

cot2

cot2

cot2CBA

(b) 2

cot2

cot2

cot4CBA

(c) 2

cot2

cot2

cotCBA

(d) 2

cot2

cot2

cot8CBA

16. sin2 18 + sin2 78 + sin2 42 = ?

(a) 1

2 (b)

3

2 (c)

2

3 (d)

5

2

17. cos2 + cos2 (60° – ) – sin2 (60° + ) = ?

(a) 2

3 (b) 1 (c)

3

2 (d)

1

2

18. 2 + sin2 (60° – ) + cos2 (30° – ) – 2

1

sec =?

(a) 3

2 (b)

5

2 (c) 1 (d)

3

2−

19. sin 12° sin 24° sin 48° sin 84°

Exercise - 1



107

(a) 1

8 (b)

7

8 (c)

1

16 (d)

9

16

20. sin2 7 + sin2 67 + sin2 53 = ?

(a) 1 (b) 1

2 (c)

5

2 (d)

3

2

21. 2

1

sin 36 – cot2 12° cot2 48° cot2 72° = ?

(a) 3 (b) 1 (c) –1 (d) 2

22. cos20 cos40 cos80 =?

(a) 1/2 (b) 1/4 (c) 1/6 (d) 1/8

23. sin2 29° + sin2 31° + sin2 89° + sin2 45° =?

(a) 3/2 (b) 1/4 (c) 2 (d) 1/8

24. If =++ CBA and ,coscoscos CBA = then CB tantan

is equal to

(a) 2

1 (b) 2 (c) 1 (d)

2

1−

25. If A, B, C, D are the angles of a cyclic quadrilateral, then

cos A + cos B + cos C + cos D =?

(a) 4 (b) 1 (c) 0 (d) –1

26. If 2 2 2cos cos cos 1A B C+ + = then, ABC is

(a) Equilateral (b) isosceles

(c) Right angled (d) none of these


108

1. 15

16cos

15

8cos

15

4cos

15

2cos

=

(a) 1/2 (b) 1/4 (c) 1/8 (d) 1/16

2. )120(cos)120(coscos 222 −+++ =?

(a) 3/2 (b) 1 (c) 1/2 (d) 0

3. The value of 12

5cos

4cos

12cos 222

++ is

(a) 2

3 (b) 3

2 (c) 2

33 + (d)

33

2

+

4. The value of 16

7sin

16

5sin

16

3sin

16sin

is

(a) 16

1 (b)

16

2 (c)

8

1 (d)

8

2

5. ,, are real numbers satisfying =++ . The

minimum value of sinsinsin ++ is

(a) Zero (b) – 3 (c) Positive (d) Negative

6. If A, B, C, D are the angles of a cyclic quadrilateral, then

=+++ DCBA coscoscoscos

(a) 2(cos cos )B C+ (b) )cos(cos2 BA +

(c) )cos(cos2 DA + (d) 0

7. If ABCD is a cyclic quadrilateral, then the value of

=−+− DCBA coscoscoscos ?

(a) 0 (b) 1

(c) )cos(cos2 DB − (d) )cos(cos2 CA −

8. If CBA coscoscos = and ,=++ CBA then the value of

CB cotcot is

(a) 1 (b) 2 (c) 3

1 (d)

2

1

9. If ,180oCBA =++ then the value of )cot(cot CB +

)cot(cot)cot(cot BAAC ++ will be

(a) CBA secsecsec (b) CBA coseccoseccosec

(c) CBA tantantan (d) 1

10. If ,180oCBA =++ then =2

tan2

tanBA

(a) 0 (b) 1 (c) 2 (d) 3

11. If )0,,( =++ CBACBA and the angle C is obtuse then

(a) 1tantan BA (b) 1tantan BA

(c) 1tantan =BA (d) None of these

12. If A, B, C are acute positive angles such that

=++ CBA and ,cotcotcot KCBA = then

(a) 33

1K (b)

33

1K (c)

9

1K (d)

3

1K

13. If ,2

3=++ CBA then =++ CBA 2cos2cos2cos

(a) CBA coscoscos41 − (b) CBA sinsinsin4

(c) CBA coscoscos21 + (d) CBA sinsinsin41 −

14. In a ABC, tan A + tan B + tan C = 6 and tan A tan B = 2

then find the value of tan C.

(a) 1 (b) 2 (c) 3 (d) –3

15. In a ABC if cos2 A + cos2 B + cos2 C = 1, then ABC is

(a) equilateral (b) isosceles

(c) right angled (d) N.O.T.

16. sin 6° sin 42° sin 66° sin 78° =?

(a) 1

8 (b)

1

16 (c)

1

16 2 (d)

1

8 2

17. If A + B + C = , then sin2 2sin

2

A+ 2sin

2

B−

2

C=?

(a) 1 – 2 cos cos sin2 2 2

A B C

(b) 1 – 2 cos cos sinA B C

(c) 1 – 2 cos2 2 2cos sin2 2 2

A B C

(d) None of these

Exercise - 2



109

18. 3 5 7

sin sin sin sin16 16 16 16

(a) 1

8 (b)

1

16 (c)

1

16 2 (d)

1

8 2

19. 2 3 4


=?

(a) 1

8 (b)

5

16 (c)

1

16 (d)

3

8−

20. If ,40cos20cos10cos =x then the value of x is

(a) 10tan4

1 (b) 10cot

8

1

(c) 10cosec8

1 (d) 10sec

8

1

21. 2 4

cos cos cos7 7 7

=?

(a) 0 (b) 2

1 (c)

4

1 (d)

8

1−

22. 2 4 8


=?

(a) 1/16 (b) 0 (c) – 1/8 (d) –1/16

23. If A, B and C are the angles of a plain triangle and

3

2

2tan,

3

1

2tan ==

BA then 2

tanC

=?

(a) 7/9 (b) 2/9 (c) 1/3 (d) 2/3

24. The value of tan 6° tan42° tan 66° tan 78° is

(a)1 (b) 1

2 (c)

1

4 (d)

1

8

25. Given that a, b, c are the sides of a triangle ABC which is

right angled at C, the then minimum value of 2

c c

a b

+

is

(a) 0 (b) 4 (c) 6 (d) 8


110

Answer key

1 a 2 b 3 a 4 a 5 a

6 c 7 d 8 a 9 c 10 a

11 b 12 a 13 b 14 d 15 c

16 b 17 d 18 b 19 c 20 d

21 b 22 d 23 c 24 b 25 c

26 c

1. (a)

tan 20° tan 80° cot 50°

= tan 20° tan 80° tan 40°

= tan (3 × 20) = tan 60° = 3

2. (b)

1 – sin 10° sin 50° sin 70°

= 1 – 1

4. sin (3 × 10) =

1 1 71 .

4 2 8− =

3. (a)

cos 20° cos 40° cos 60° cos 80°

= 1 1

2 4 cos (3 × 20) =

1 1 1 1

2 4 2 16 =

4. (a)

1 1

cos15 cos7 cos822 2

= 1 1

cos15 cos7 sin72 2

= 1 1 1cos15 2cos7 sin7

2 2 2

= 2cos15 sin15

2 2

= ( )1 1 1 1

sin 2 154 4 2 8

= =

5. (a)

tan6 tan66 tan54 tan42 tan78

tan54

= ( )tan 3 6 tan42 tan78

tan54

=

( )tan 3 18 tan541

tan54 tan54

= =

6. (c)

sin 20° sin 40° sin 60° sin 80°

sin 60° sin 20° sin 40° sin 80°

Using sin sin (60° – ) sin (60° + ) = 1

4sin 3

sin 60° 1

4 sin (3 × 20) =

3 1 3 3

2 4 2 16 =

7. (d)

sin12 sin48 sin72 sin54

sin72

=

( )sin 3 12 sin541

4 sin72

= sin36 sin54

4sin72

=

1 sin36 cos36

4 2sin36 cos36

=

1

8

8. (a)

cot 15° cot 25° cot 35° cot 85° = cot 15° cot (3 × 250)

= cot 15° cot 75° = cot 15° tan 15° = 1

9. (c)

cos 10° cos 30° cos 50° cos 70°

cos 30° cos 10° cos 50° cos 70°

Using cos cos (60° – ) cos (60° + ) = 1

4 cos 3

= cos 30° × 1

4 cos (3 × 10) =

3 1 3 3

2 4 2 16 =

10. (a)

sin 6° cos 48° cos 24° cos 12°

sin 6° cos 12° cos 24° cos 48°

Using cos cos 2 cos 4 …… cos2n-1

= n

n

sin2

2 sin

= sin 6°.

( )3

3

sin2 12

2 sin12 =

sin6 .sin96

8.sin12

= 2 sin6 cos6 sin12 1

16 sin12 16 sin12 16

= =

11. (b)

( )cos2 cos2 cos2x y z+ −

= 22cos( )cos( ) (2cos 1)x y x y z+ − − −

1cos2)cos()cos(2 2 +−−+= zyxyx

1cos2)cos()cos(2 2 +−−−= zyxz

= 22cos cos( ) 2cos 1z x y z− − − +

=1 2cos {cos( ) cos }z x y z− − +

Solutions

Exercise - 1



111

)}cos(){cos(cos21 yxyxz +−−−=

( )1 2cos 2sin sinz x y= − = 1 4sin sin cosx y z−

12. (a)

2 + + = , 2 2 2

+ + =

tan2 2 2

+ +

= tan = 0

tan2 2 2

+ + =

tan tan tan tan tan tan2 2 2 2 2 2

1 tan tan tan tan tan tan2 2

2 2 2 2

+ + −

− − −

tan tan tan tan tan tan

2 2 2 2 2 2 0

1 tan tan tan tan tan tan2 2 2 2 2 2

+ + −

=

− − −

tan tan tan2 2 2

+ + - tan tan tan

2 2 2

= 0

2

tan2

tan2

tan2

tan2

tan2

tan

=++

13. (b)

CBA 222 sinsinsin ++

CBA 222 sincos1cos1 +−+−=

)cos()cos(cos2 2 CBCBA −+−−=

)]cos([coscos2 CBAA −−−=

)]cos()cos([cos2 CBCBA −−+−−=

( )2 cos 2cos cosA B C= +

2coscoscos2sinsinsin 222 =−++ CBACBA

14. (d)

CBA 2sin2sin2sin −+

= )sin()cos(2cossin2 CBCBAA −++

),cos()cos(,,{ ACBACBCBA −=+−=+=++

}sin)sin(,cos)cos( ACBACB =+−=+

= 2sin cos 2cos sin( )A A A B C− −

)]sin([sincos2 CBAA −−=

)]sin()[sin(cos2 CBCBA −−+=

2cos 2cos sinA B C= 4cos cos sinA B C=

15. (c)

oCBA 180=++

2

9022

CBA o −=+

−=

+

290cot

22cot

CBA o

cot cot 1

12 2 tan2

cot cot cot2 2 2

A B

C

B A C

−

= =

+

2

cot2

cot2

cot12

cot2

cotABCBA

+=

−

cot cot cot cot cot2 2 2 2 2

A B C C B= + cot

2

A+

16. (b)

Value of cos2 + cos2 (60° – ) + cos2 (60° + ) is always

3

2 irrespective of the value of .

So required answer is 3

2

17. (d)

cos2 + cos2 (60° – ) – (1 – sin2 (60° + ))

cos2 + cos2 (60° – ) + cos2 (60° + ) – 1 = 3 1

12 2

− =

18. (b)

1 + 1 – 2

1

sec + sin2 (60° – ) + cos2 (30° – )

= 1 + 1 – cos2 + sin2(60°– ) + cos2 [900– (60 + )]

= 1 + sin2 + sin2 (60° – ) + sin2 (60° + )

= 1 + 3 5

2 2=

19. (c)

sin12 sin48 sin72 sin24 sin84

sin72

= ( )sin 3 12 sin24 sin841

4 sin72

= 1 sin24 sin36 sin84

4 sin72

=

( )sin 3 241 1

4 4 sin72 16

=

20. (d)

Value of sin2 + sin2 (60° – ) + sin2 (60° + ) is always

3

2 irrespective of the value of . So required answer is

3

2.

21. (b)

cosec2 36° – (cot 12°. cot 48°. cot 72°)2


112

= cosec2 36° – cot2 (3 × 12) = cosec2 36° – cot2 36 = 1

22. (d)

cos20 cos40 cos80o o o

= cos20O cos (60O – 20O) cos( 60O + 200)

= 1

cos(3 20 )4

= 1

cos604

= 1

8

23. (c)

sin2 29° + sin2 31° + sin2 89° + sin2 45°

= sin2 29° + sin2 (60° – 29°) + sin2 (60° + 29°) + sin2 45°

= 3

2 + sin2 45° =

3 1 4

2 2 2+ = = 2

24. (b)

cos cos cosA B C=

CBCB coscos)](cos[ =+−

CBCB coscos)cos( =+−

CBCBCB coscos]sinsincos[cos =−−

CBCB coscos2sinsin = 2tantan =CB

25. (c)

ABCD is a cyclic quadrilateral

A + C = 180°, B + D = 180°

C = 180° – A, D = 180° – B

cos C = cos (180° – A) = –cos A, cos D = cos (180° – B)

= – cos B

cos A + cos C = 0, cos B + cos D = 0

cos A + cos B + cos C + cos D = 0

26. (c)

2 2 2cos cos cos 1A B C+ + =

cos2A + cos2B – (1 – cos2C) = 0

cos2A + cos2B – sin2C = 0

cos2A + cos (B + C) cos (B – C) = 0

2 cos A cos B cos C = 0

Hence, either A or B or C is 90°

Answer key

1 d 2 a 3 a 4 b 5 c

6 d 7 a 8 d 9 b 10 b

11 b 12 a 13 d 14 c 15 c

16 b 17 a 18 d 19 b 20 b

21 d 22 d 23 a 24 a 25 d

1. (d)

15

16cos

15

8cos

15

4cos

15

2cos

4

4

2sin 2

152

2 sin15

= =

32sin

152

16sin15

=

sin 215

216sin

15

+

=

2sin

1 115216 16

sin15

=

2. (a)

)120(cos)120(coscos 222 oo −+++

22 )120(cos)120(coscos oo −+++=

)120(cos)120(cos2 oo −+−

= 22 2 2cos 2cos cos120 2 cos sin 120o o + − −

o120sin2cos2coscos 2222 +−+=

2

3

4

32120sin2 2 === o

3. (a)

12

5cos

4cos

12cos 222

++

+

+

−=

12

5cos

2

1

12sin1 2

2

2

−++=

12sin

12

5cos

2

11 22

3 5 5

cos cos2 12 12 12 12

= + + −

3

cos cos2 2 3

= +

3 1 30

2 2 2= + =

4. (b)

3 5 7


Exercise - 1



113

1 3 5 7

2sin sin 2sin sin4 16 16 16 16

=

−

−=

4

3cos

8cos

4cos

8cos

4

1

+

−=

2

1

8cos

2

1

8cos

4

1

−=

−= 1

8cos2

8

1

2

1

8cos

4

1 22

16

2

2

1

8

1

4cos

8

1==

=

5. (c)

When =++

)sin(sinsinsin ++++

0sinsinsin ++

The minimum value of sinsinsin ++ is always

positive.

6. (d)


So CACA −==+ 180180

CCA cos)180cos(cos −=−=

0coscos =+ CA .....(i)

Similarly, 0coscos =+ DB ..... (ii)

Adding, + + + =cos cos cos cos 0A B C D

7. (a)


+ = 180A C CA −= 180

CCA cos)180cos(cos −=−=

0coscos =+ CA .....(i)

Now ,180=+ DB then 0coscos =+ DB ....(ii)

Subtracting (ii) from (i), we get

0coscoscoscos =−+− DCBA

8. (d)

CBA coscoscos =

ACBCBA −=+=++

cos( ) cos( )B C A + = − cos( ) cosB C A+ = −

cos cos sin sin cos cosB C B C B C− = −

( Given cos cos cos )A B C=

CBCB sinsincoscos2 =

cos cos 1

sin sin 2

B C

B C =

1cot cot

2B C =

9. (b)

cot cotB C+

= sin cos sin cos

sin sin

C B B C

B C

+

CB

CB

sinsin

)sin( +=

CB

Ao

sinsin

)180sin( −=

CB

A

sinsin

sin=

Similarly, AC

BAC

sinsin

sincotcot =+

and BA

CBA

sinsin

sincotcot =+

Now,

)cot)(cotcot)(cotcot(cot BAACCB +++

sin sin sin

sin sin sin sin sin sin

A B C

B C C A A B=

=cos cos cosecA ecB ecC

10. (b)

oCBA 180=++

+−=

222

CBA

+=

22tan

2cot

CBA

2tan

2tan1

2tan

2tan

2tan

1

CB

CB

A−

+

=

1 tan tan tan tan tan tan2 2 2 2 2 2

B C A B A C− = +

tan tan tan tan tan tan 12 2 2 2 2 2

A B B C A C+ + =

i.e. , = 12

tan2

tanBA

11. (b)

CBACBA −=+=++ )tan()tan( CBA −=+

)tan(tantan1

tantanC

CA

BA−=

−

+ C

BA

BAtan

tantan1

tantan−=

−

+

Now C is an obtuse angle,

hence

0tan0tan − CC

0tantan10tantan1

tantan−

−

+ BA

BA

BA

BA,( are acute angles; 0tan,0tan BA )

1tantan BA

12. (a)

=++ CBA

CBACBA tantantantantantan =++


114

Now A.M. G.M.

3/1)tantan(tan

3

tantantanCBA

CBA

++

3/1)tantan(tan3

tantantanCBA

CBA

3)tantan(tan 3/2 CBA

33

13

13

1 2/3

3/2

K

KK

13. (d)

CBA 2cos2cos2cos ++

CBABA 2cos)cos()cos(2 +−+=

CBAC 2cos)cos(2

3cos2 +−

−=

CBAC 2sin21)cos(sin2 −+−−=

}sin){cos(sin21 CBAC +−−=

+−+−−= )(

2

3sin)cos(sin21 BABAC

)}cos(){cos(sin21 BABAC +−−−=

CBA sinsinsin41 −= .

14. (c)

In a ABC we know that

A + B + C = and

tan A + tan B + tan C = tan A tan B tan C

6 = 2 tan C

tan C = 3

15. (c)

cos2 A + cos2 B + 1 – sin2 C = 1

cos2 A + cos2 B – sin2 C = 0

cos2 A + cos (B + C) cos (B – c) = 0

cos2 A – cos A cos (B – C) = 0

( A + B + C = B + C = – A)

cos2 A = cos A cos (B – C)

cos A = cos (B – C) A = B – C A + C = B

We know that A + B + C = 2B =

B = 2

Hence is right angled.

16. (b)

Ist method

cos 84° cos 48° cos 24° cos 12°

= 2sin96

cos842 8sin12

=( ) ( )sin 96 84 sin 96 84

16sin12

+ + −

= sin180 sin12 1

16sin12 16

+ =

IInd method:

= sin6 sin66 sin54 sin18 sin42 sin78

sin54 sin18

=

sin18 sin5414 4

sin54 sin18 16

=

17. (a)

sin2 2sin

2

A+ 2sin

2

B−

2

C

= sin sin2 2

A C A C+ −

+ 1-2cos

2

B

= cos 2sin cos 12 2 2

B A C B− − +

= cos sin cos 12 2 2

B A C B − − +

= cos sin sin 12 2 2

B A C A C − + − +

= 1 – 2 cos cos sin2 2 2

A B C

18. (d)

3 5 7sin sin sin sin

16 16 16 16

= 3 3sin sin sin sin

16 16 2 16 2 16

− −

= 3 3

sin sin cos cos16 16 16 16

=

1 3sin sin

4 8 8

= 1sin sin

4 8 2 8

−

=1

sin cos4 8 8

=

1 1sin

4 2 4

=

1

8 2

19. (b)

2 3 4


= 2 2sin sin sin sin

5 5 5 5

− −

=

22

sin sin5 5

=

21 2

2sin sin2 5 5

=

2

1 2 2cos cos

4 5 5 5 5

− − +

=

21 3

cos cos4 5 5

−

[∵ cos(-) = cos ]



115

= ( )21

cos36 cos1084

− =

2

1 5 1 5 1

4 4 4

+ −+

=

2

1 5 5

4 2 16

=

20. (b)

ooox 40cos20cos10cos=

1

[2 sin10 cos10 cos 20 cos 40 ]2 sin10

o o o o

o=

]40cos20cos20sin2[10sin2.2

1 ooo

o=

1

[2sin40 cos40 )2.4sin10

o o

o= =

1(sin80 )

8sin10o

o

oo

o10cot

8

110cos

10sin8

1==

21. (d)

2 4

cos cos cos7 7 7

3

3

sin 2 .7

2 sin7

=

=

8sin

7

8sin7

=

sin7

8sin7

+

=

sin17

88sin

7

−

= −

22. (d)

5

8cos

5

4cos

5

2cos

5cos

=

4

4

2sin

5

2 sin5

=

16sin

5

16sin5

=

sin 35

16sin5

+

= sin

5

16sin5

−

= 1

16−

23. (a)

=++ CBA

−=

+

22tan

2tan

CBA

tan tan

2 2 cot2

1 tan .tan2 2

A BC

A B

+

=

−

1 293 3 cot

1 2 7 21

3 3

C+

= =

−

9

7

2tan =

C

24. (a)

tan 6° tan 42° tan 66° tan 78°

= tan6° tan(60° – 18°) tan(60° + 6°) tan(60° + 18°)

= ( ) ( ) ( )tan6 tan 60 6 tan18 tan 60 18 tan 60 18

tan18

+ − +

= ( ) ( )tan6 tan 60 6 tan 3 18

tan18

+

= ( ) ( )tan6 tan 60 6 tan 60 6 tan18

1tan18 tan18

− + = =

25. (d)

A b C

a c

B

a = c sin , b = c cos

2

c c

a b

+

= 2

1 1

sin cos

+

= ( )

2

4 1 sin2

sin 2

+

= 2

1 14

sin 2 sin2

+

where 0 < <

2

2

min

c c

a b

+

= 8 when 2 = 90° = 45°


116

(i) sin

Max. value = +1

Min value = -1

From 0ᵒ to 90ᵒ sin curve increasing

(ii) cos

Max value = +1

Min. value = -1

From 0ᵒ to 90ᵒ cos curve decreasing

(iii) tan

Min. value = - ∞

Max. value = +∞

From 0ᵒ to 90ᵒ tan curve increasing

(iv) cot

Min. value = - ∞

Max. value = + ∞

From 0ᵒ to 90ᵒ cot curve decreasing

(v) sec

Range (-∞, - 1] ⋃ [1, ∞)

(vi) cosec

Range (-∞, - 1] ⋃ [1, ∞)

♦ 0 < A, B < 90ᵒ

and A > B

then sinA > sinB

cosA < cosB

and tanA > tanB

------------------* * *----------------

Graphs of Trigonometric Ratios 10



117

Maximum and minimum value of (a sin + b cos)

a sin + b cos

= 2 2

2 2 2 2

a ba b sin cos

a b a b

+ +

+ +

= 2 2a b+ [sin sin + cos cos]

= 2 2a b+ cos ( – ) ⇒– 1 cos ( – ) 1

( )2 2 2 2 2 2a b a b cos a b− + + − +

* Maximum value of sinn cosn is

n1

2

** * Inverse ratio

y = a tan2 x + b cot2 x Where a, b > 0

y= ( )2

a tanx b tanx 2 ab− +

as a tan x = b cot x tan2x = b

a (possible)

ymin = 2 ab

( ) y 2 ab,

Minimum value of a tan2 x + b cot2 x = 2 ab

* y = a cosec2 x + b sin2 x

Minimum value = 2 ab ⇒ ( ) y 2 ab,

* y = a sec2 x + b cos2 x

Minimum value = 2 ab

( ) y 2 ab,

* Minimum value of a sec2 + b cosec2

a sec2 + b cosec2

= a(1+tan2) + b(1+cot2)

= a + b + a tan2 + b cot2

Min value = a + b + 2 ab = ( )2

a b+


118

1. why sin 1 is greater than sin1ᵒ?

Sol. sin1 = sin 57ᵒ.16’

As sin is increasing in 0,2

sin 57ᵒ 16’ > sin1ᵒ

2. which of the following is true?

(a) sin1 > sin1ᵒ (b) cos 1 < cos1ᵒ

(c) tan 1 > tan1ᵒ (d) all of these

Sol. (a) sin1 >sin1ᵒ above explained

(b) cos1 = cos 57ᵒ16’

as in 0,2

cos is decreasing

∴ cos57ᵒ16’ < cos1ᵒ

tan1 = tan57ᵒ16’

tan is increasing in 0,2

∴ tan 1 > tan1ᵒ

3 Which of the following is possible?

(a) sin = 4

3 (b) tan = 102

(c) sec = 3

4 (d) cos =

2

2

1

1

p

p

+

−(P≠±1)

Sol. sin = 4

3 is not possible as -1≤ sin≤1

tan = 102 is possible as tan can take any real value

sec =3

4 is not possible as sec ⋲ (-∞, -1]⋃[1, ∞)

cos = 2

2

1

1

p

p

+

− is not possible, as in

2

2

1

1

p

p

+

− numerator is

always greater than the denominator for any value of p

other than p = 0 hence 2

2

1

1

p

p

+

− does not lie in [-1, 1]

4 Find maximum and minimum value of sinx + cos x.

Sol. y = sin x + cos x

Max. value = 1 1 2+ =

Min. value = 1 1 2− + = −

5 y = 3 sin x + 4 cos x

Max. = 2 23 4+ = 5

Min. = 2 23 4− + = – 5

6 y = 5 sin x – 12 cos x + 40

Max. Value = 2 25 12+ + 10

= 13 + 10 = 23

Min. Value = 2 25 12− + + 10

= – 13 + 10 = – 3

7 Maximum value of (2 sin + 3 cos ) is.

(a) 2 (b) 13 (c) 15 (d) 1

Sol. (b)

y = 2 sin + 3 cos

max. value = 2 22 3 13+ + =

8 Find the minimum value of (24 sin + 7 cos ) is.

(a) –7 (b) 17 (c) –24 (d) - 25

Sol. (d)

y = 24 sin + 7 cos

min value = 2 224 7− + = - 25

9 Maximum and minimum values of 3 sin – 5 cos is.

(a) 3, –3 (b) 5, –5 (c) 34, 34− (d) 4, –4

Sol. (c)

y = 3 sin – 5 cos

max. value = 2 23 5 34+ + =

min. value = 2 23 5 34− + =

10 Find least value of 5 cos +9.

(a) 5 (b) 7 (c) –7 (d) –5

Sol. (b)

–1 cos 1 –5 5 cos 5

–5 + 12 5 cos + 12 5 + 12

7 5 cos + 12 17

Least value = 7

Or

Least value of cos = –1

Least value of 5 cos + 12 = – 5 + 12 = 7

11 Find least value of 4 sin – 3 cos + 5

Examples



119

(a) 0 (b) 4 (c) 3 (d) 5

Sol. (a)

f() = 4 sin – 3 cos + 5

Least value of 4 sin – 3 cos = 2 24 3− + = – 5

Least value of f() = – 5 + 5 = 0

12 y = 8 sec2 x + 18 cos2 x

ymin = 2 8 18 = 2 × 12 = 24

13 The least value of a cosec2 + 16 sin2 is.

Sol. Least value = 2 9 16 = 2 × 12 = 24

14 The least value of 25 tan2 + 9 cot2 is.

Sol. Least value = 2 25 9 = 2 × 5 × 3 = 30

15 Minimum value of 49 sec2 + 9 cosec2 = ?

(a) 100 (b) 42 (c) 49 (d) 9

Sol. (a)

y = 49 sec2 + 9 cosec2

Minimum value = ( )2

49 9+ = (7 + 3)2 = 100

16 The least value of sin4 + cos4

(a) 1

4 (b) 3 (c)

1

2 (d)

3

4

Sol. (c)

sin4 + cos4 = 1 – 2 sin2 cos2 = 1 – 2sin 2

2

Value will be least when value of 2sin 2

2

is maximum.

max.

2sin 2

2 = ½, when = 450

Now max. value of sin4 + cos4 = − =


120



121

1. Which is smaller, sin 67° or cos 67°?

(a) sin 67° (b) cos 67°

(c) Both are equal (d) None of these

2. If the sine of an angle is 2/5, then cosine of that angle is

(a) equal to 2/5 (b) less than 2/5

(c) greater than 2/5 (d) cannot be determined

3. The value of sincos ba + lies between

(a) ba − and ba + (b) a and b

(c) )( 22 ba +− and )( 22 ba + (d) 22 ba +− and 22 ba +

4. The maximum value of sin4cos3 − is

(a) 3 (b) 4 (c) 5 (d) None of these

5. The value of x, for maximum value of (sinx + cosx) is:

(a) 30° (b) 45° (c) 60° (d) 90°

6. Minimum value of 22 cos4sin5 + is

(a) 1 (b) 2 (c) 3 (d) 4

7. If 0 < < 2

, which of the following is true?

(a) sin2 + 2

1

sin < 2 (b) sin2 +

2

1

sin = 2

(c) sin2 + 2

1

sin > 2 (d) None of these

8. The maximum value of

+−

− xx

3cos

3cos 22

is

(a) 2

3− (b)

2

1 (c)

2

3 (d)

2

3

9. 22 cottan + is

(a) 2 (b) 2 (c) 2− (d) None of these

10. The value of x for the maximum value of

xx sincos3 + is

(a) 30° (b) 45° (c) 60° (d) 90°

11. The minimum value of 5sin4cos3 ++ xx is

(a) 5 (b) 9 (c) 7 (d) 0

12. The greatest and least value of xx cossin are

(a) 1,1 − (b) 2

1,

2

1− (c)

4

1,

4

1− (d) 2,2 −

13. The maximum value of xx 22 cos3sin4 + is

(a) 3 (b) 4 (c) 5 (d) 7

14. The maximum value of

++

+

6cos

6sin

xx in the

interval

2,0

is attained at

(a) 12

=x (b)

6

=x (c)

3

=x (d)

2

=x

15. The minimum value of 22 cot4tan9 + is

(a) 13 (b) 9 (c) 6 (d) 12

16. If ,sincos 42 +=A then for all values of

(a) 21 A (b) 116/13 A

(c) 16/134/3 A (d) 14/3 A

17. If ,cossin 42 +=A then for all real values of

(a) 21 A (b) 14

3 A

(c) 116

13 A (d)

16

13

4

3 A

18. What is the maximum value of sin6 + cos6 ?

(a) 1

2 (b)

1

4 (c) 1 (d) None of these

19. The value of x for maximum value of )cossin3( xx + is

(a) 30o (b) 45o (c) 60o (d) 90o

20. The least value of sin2 + cosec2 + cos2 + sec2 is:

(a) 3 (b) 4 (c) 5 (d) 6

21. The greatest value of 16sin x 8cos x is:

(a) 35 (b) 34 (c) 3 (d) 33

22. If O° < A < 90° and cos A – sin A > 0 then cos A + sin A

cannot be greater than:

(a) 1

3 (b)

1

2 (c)

1

2 (d) 2

Exercise - 1


122

23. Let P = sin (sin + sin3 ). The P

(a) O only when O (b) O for all real

(c) O for all real (d) O only when O

24. The ratio of the greatest value of 2 – cos x + sin2 x to its

least value is:

(a) 1

4 (b)

9

4 (c)

13

4 (d)

7

4

25. The least value of sin2 + cos2 + tan2 + cot2 +

sec2 + cosec2 is:

(a) 8 (b) 7 (c) 5 (d) 6



123

Answer key

1 b 2 c 3 d 4 c 5 b

6 d 7 c 8 c 9 a 10 a

11 d 12 b 13 b 14 a 15 d

16 d 17 b 18 b 19 c 20 c

21 a 22 d 23 c 24 c 25 b

1. (b)

cos 67° = cos (90° – 23°) = sin 23 °

In 1st quadrant as increases, the value of sin

increases.

sin 67° > sin 23°,

hence cos 67° is smaller.

2. (c)

cos = − = − = 2 4 21 41 sin 1

25 25 25

So, cos > 2

5

3. (d)

cos sina b + = 2 2a b+

++

+ 2222

sincos

ba

b

ba

a

= 2 2 sin( )a b + +

Since, ,1)sin(1– +

Then 2222 )sin( baba +++−

4. (c)

the maximum value of cossin ba + is 22 ba ++ and

minimum value is 22 ba +−

the maximum value is

(3cos 4sin ) + = 2 23 ( 4)+ − = 5

and the minimum value = - 5

5. (b)

sin x + cos x = 1 1

2 sinx cosx2 2

+

= 2 (sin x cos 45° + cos x sin 45°)

= 2 sin (x + 45°) = 2 , when x = 45°

Max. (sin x + cos x) = 2

6. (d)

Let y = 2 25sin 4cos + = 24 sin +

04)( + f )0sin( 2

The minimum value of )(f is 4

7. (c)

sin2 + 2

1

sin = sin2 +

2

1

sin – 2 + 2

=

21

sinsin

−

+ 2 > 2

0 0 sin 12

8. (c)

+−

− xx

3cos

3cos 22

cos cos3 3

x x

= − + + cos cos

3 3x x

− − +

= 2cos cos 2sin sin3 3

x x

= 2

sin sin 23

x

=3

sin 22

x

So maximum value is3

2 { 1 sin 2 1}x−

9. (a)

We know that2

10x

x

−

, 2

2

12 0x

x+ −

Put x = tan , 2

2

1tan 2

tan

+ 2 2tan cot 2 +

10. (a)

Let xxxf sincos3)( +=

+=

+=

3sin2sin

2

1cos

2

32)(

xxxxf

But 13

sin1

+−

x

Hence, )(xf is maximum, if 903

x

+ = 30x =

11. (d)

The minimum value of xx sin4cos3 + is 2 23 4− + = -

5

Solutions


124

Hence the minimum value of 5sin4cos3 ++ xx

055 =+−=

12. (b)

Let y = sin cosx x = 1

sin 22

x

We know 1 sin2 1x− 1 1 1

sin 22 2 2

x−

Thus the greatest and least value of )(xf are 2

1 and

2

1− respectively

13. (b)

3sincos3sin4)( 222 ++= xxxxf and 1|sin|0 x

Maximum value of 3sin2 +x is 4

14. (a)

2 cos6 4

x

+ − = 2 cos

12x

−

Hence maximum value will be at 12

=x

15. (d)

the minimum value = 2 9 4

A.M. G.M.

2 29 tan 4cot

2

+ 2 24cot 9tan

2 2tan 4cot 12 +

Therefore, the minimum value is 12.

16. (d)

42 sincos +=A 222 sin.sincos +=A

22 sincos +A ]1sin[ 2

1A

Again 4242 sin)sin1(sincos +−=+=A

4

3

4

3

2

1sin

2

2 +

−= A

Hence, 14/3 A

17. (b)

We have 42 cossin +=A

22222 cossincoscossin ++=

(since )1cos2

11cossin 42 + A

Again, 4242 coscos1cossin +−=+

4

3

4

3

2

1cos1coscos

2

224 +

−=+−=

Hence, .14

3 A

18. (b)

x = sin6 + cos6 x = (sin2 )3 + (cos2 )3

x = (sin2 +cos2 )(sin4 + cos4 – sin2 cos2 )

x = 1 × [(sin2 + cos2 )2 – 3 sin2 cos2 ]

x = 1 – 3 sin2 cos2

x = 1 – 3

4 (2sin cos )2

x = 1 – 3

4 (sin 2)2 = 1 –

3

4sin 2

0 sin2 2 1

at sin2 2 = 0

x = 1 – 3

4 (0) = 1 and at sin2 2 = 1

x = 1 – 3

4 (1) =

1

4

i.e. 1

4 x 1

i.e. least value of sin6 + cos6 = 1

4

19. (c)

The greatest value of xx cossin3 + is 213 =+ and

obviously it will be at = 60x .

Alter :

+=

+

6sin2cos

2

1sin

2

32

xxx

As xsin is maximum at ,2

=x so

26

=+x or

3

=x

20. (c)

(sin2 + cos2 ) + (cosec2 + sec2 )

= 1 + (cosec2 + sec2 )

Minimum value of

(a cosec2 + b sec2 ) = ( )2

a b+

Minimum value of

(cosec2 + sec2 ) = ( )2

1 1+ = 4

min value = 4 + 1 = 5

21. (a)

16sin x8cos x = 24 sin x23 cos x = 24 sin x + 3 cos x

For maximum value,

4 sin x + 3 cos x must be maximum and maximum

value of:

4 sin x + 3 cos x = 2 24 3+ = 5

Greatest value of

16sin x8cos x = 25 = 32



125

22. (d)

cos A – sin A > 0

sin A < cos A

tan A < 1 A < 45°

sin A + cos A < sin 45° + cos 45°

< +1 1

2 2

< 2

i.e. sin A + cos A cannot be greater than 2 .

23. (c)

P = sin (sin + sin3)

= sin (2 sin 2cos ) = sin 2 (2 sin .cos )

= sin 2 sin 2 = sin2 2 P = sin2 2 0

For all real

24. (c)

y = 2 – cos x + sin2 x = 2 – cos x + 1 – cos2 x

= – (cos2 x + cos x) + 3 =

21 1

cosx2 4

+ −

+ 3

=

213 1

cosx4 2

− +

Max. value of y occurs at cos x = 1

2− and it is

13

4

and min. value occurs at cos x = 1 and it is 1

The requited ratio is 13

4

25. (b)

sin2 + cos2 + tan2 + cot2 + sec2 + cosec2

= (sin2 + cos2 ) + tan2 + cot2 + (1 + tan2 ) + (1 +

cot2 )

= + 2(tan2 + cot2 )

Minimum value of

(a tan2 + b cot2 ) = 2 ab

Minimum value of

tan2 + cot2 = 2

min value = 3 + 2 × 2 = 7

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