chapter · 2020. 9. 26. · chapter 4 section - a angle and its units (1) angle consider a ray oa....

4CHAPTERSECTION - A

ANGLE AND ITS UNITS(1) AngleConsider a ray OA

. If this rayrotates about its end points Oand takes the position OB, thenthe angle AOB has beengenerated.An angle is considered as the

= angleO A

B

Initial sidefigure obtained by rotating agiven ray about its end-point.

The initial position OA is called the initial side andthe final position OB is called terminal side of the angle.The end point O about which the ray rotates is called thevertex of the angle.Sense of an Angle :The sence of an angle is said to be positive or negativeaccording as the initial side rotates in anticlockwise orclockwise direction to get to the terminal side.

= +ve O A

B

Anticlockwise direction

= –veO A

B

Clockwise direction

Some Useful termsQuadrantLet XOX and YOY be two lines at right angles in theplane of the paper. These lines divide the plane of the paperinto four equal parts which are known asquadrants.The lines XOX and YOY are known as x-axisand y-axis respectively.These two lines taken togetherare known as regions XOY,YOX, XOY and YOX are

II-quadrant I-quadrant

III -quadrant IV-quadrant

Y

X X

Y

O

known as the first, the second,the third and the fourthquadrant respectively.

Angle In Standard Position :An angle is said to be in standard position if its vertex co-incides with the origin O and the initial side coincides withOX i.e. the positive direction of x-axis.

Co–terminal Angles :Two angles with different measures but having the sameinitial sides and the same terminal sides are known as co-terminal angles.

X

P P

MM O X

Y

Y

(2) SYSTEM OF MEASUREMENT OF ANGLEThere are three system for measuring angles.

(i) Sexagesimal or English system(ii) Centesimal or French system(iii) Circular system

Some Important Conversion : Radian = 180º

One radian = º180

Relation between systems of measurement of angles

D G C90 100

2

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4.2 Theory and Exercise Book


SOLVED EXAMPLE

EXAMPLE 1Express 1·2 radians in degree measure.

SOLUTION

1·2 radians = 1·2 ×180π

degrees = 1·2 ×7/22

180

[ =227

(approx).]

=1·2×180× 7

22 = 68·7272 = 68º (·7272 × 60)’ = 68º (43·63)’

= 68º 43’ (·63 × 60)” = 68º 43’ 37·8”

EXAMPLE 2Express 45°20'10" in rad measure ( = 3.1514)

SOLUTION

10" = min1060 =

deg rees1060 60 =

degrees1360

20' = deg rees2060 =

deg rees13

45°20'10" = degrees1 1

45360 3

=16200 1 120

360

= 16321360

Now16321360

= rad16321360 180

=.

16321 3 1416

360 180 = .51274 054

64800 = 0.79 rad

SECTION - BBASIC DEFINATION OF TRIGONOMETRIC RATIO

In the right angled triangle OMP , we have base (OM) = x,perpendicular (PM) = y and hypotenuse (OP) = r, then wedefine the following trigonometric ratios which are knownas trigonometric function.

sin = PH

yr

cos = BH

xr

P

y

MO x

r

tan = PB

yx

cot = BP

= xy

sec = HB

rx

cosec = HP

ry

EXAMPLE 3If cosecA + cotA = 11/2, then find the value of tanA

SOLUTION

cosecA + cotA = 112 ...(1)

1

cosecA cot A = 211

cosecA – cotA = 211 ...(2)

(1) – (2)

2cot A = 112 –

211 =

11722

tanA = 44

177 .

EXAMPLE 4

If 5 tan = 4, then find the value of 5 sinθ – co sθsinθ + 3cosθ

SOLUTIONtan = 4/5

5sinθ – 3cosθs inθ + 2cosθ =

5tanθ – 3tanθ + 2

=

45 × – 35

4 + 25

= 5/14

EXAMPLE 5If angle C of triangle ABC is 90°, then prove that

ctan A tan Bab

2

(where, a, b, c are sides opposite to angles

A,B,C respectively)

SOLUTIONDraw ABC with C 90

tan A + tan B = a b+b a

= 2 2 2a + b c=ab ab B

A

C

bc

a

Trigonometry Ratio 4.3

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SECTION - CSIGNS OF TRIGONOMETRIC RATIOS

II-quadrant I-quadrant

III-quadrant IV-quadrant

Y

X X

Y

S A all positive

T Ctan and cot are positive

cos and sec are positive

sin and cosec are positive

SOLVED EXAMPLE

EXAMPLE 6Which statement is true(A) tan 3 > cos 7 (B) cot 4 < cot 12(C) sin 15 < sin 11 (D) None of these

SOLUTION Dtan 3 IInd quadrant – vecos 7 Ist quadrant + vesin 27 IInd quadrant + ve

SECTION - DFUNDAMENTAL IDENTITIES

(a) sin = 1

cosec

(b) cos = 1

sec

(c) cot = 1

tancossin

(d) 1 + tan2 = sec2 or, sec2 – tan2= 1

(sec – tan) = 1

(sec tan )

(e) sin2 + cos2 = 1(f) 1 + cot2 = cosec2

(cosec – cot) = 1

cos cotec

SOLVED EXAMPLE

EXAMPLE 7Simplify the expression

22

b - a sinx1 a. a + btan xb - a b - a1+ sinx

a

where b > a > 0.

SOLUTIONAfter a few simple manipulations, this expression (forbrevity denote it by P) can be rewritten

P = 2 2

2 2 2

sinx a + btan x sinx a + btan x=a + (b - a)sin x acos x + bsin x

Some students handle this as follows:

2 2 22

2sin x acos x + bsin xa + btan x = a + b =

cosxcos xand get a wrong answer: P = tan x. In this

transformation what we actually have to simplify is the

expression 2cos x which is equal to |cos x|. And so thefinal result is P = sinx / |cos x|.

EXAMPLE 8If tan + cot = a then find the value oftan4 + cot4

SOLUTION(tan + cot)2 = a2 tan2 a + cot2a = (a2 – 2) tan4 + cot4 = a4 + 4 – 4a2 – 2= a4 – 4a2 + 2

EXAMPLE 9

Prove that 1 + sinθ1- sinθ

= sec + tan,π π- < θ <2 2 .

SOLUTION

L.H.S. = 1 + sinθ1- sinθ

= 1 + sinθ 1+ sinθ1- sinθ 1+ sinθ

= 2

2

1 + sinθ1- sin θ

= 2

2

1 + sinθcos θ

=1+ sinθ

cosθ = 1 sinθ+

cosθ cosθ = sec + tan

= R.H.S



SECTION - EREDUCTION FORMULAE

If is any angle , then , 90 ± , 180 ± , 270 ± ,360 ± etc. are called ALLIED ANGLES .

(a) sin ( ) = sin cos ( ) = cos tan ( ) = – tan

(b) sin (90°- ) = cos cos (90° ) = sin tan (90° ) = cot

(c) sin (90°+ ) = cos cos (90°+ ) = sin tan (90°+ ) = cot

(d) sin (180° ) = sin cos (180° ) = cos tan (180° ) = tan

(e) sin (180°+ ) = sin cos (180°+ ) = cos tan (180°+ ) = tan

(f) sin (270° ) = cos cos (270° ) = sin tan (270° ) = cot

(g) sin (270°+ ) = cos cos (270°+ ) = sin tan (270°+ ) = – cot

EXAMPLE 10Prove that sin(–420°)(cos390°)+cos(–660°)(sin330°)= –1

SOLUTIONL.H.S. = sin(-420°)(cos390°) + cos(–660°)(sin330°)= – sin420°cos390° + cos660°sin330°

Q sin -θ = -sinθ,cos -θ = cosθ = – (sin90°×4+60°)cos(90°×4+30°)+cos(90°×7+ 30°)

sin(90° × 3 + 60°)= – (sin60°)(cos30°) + ( sin30°)(–cos60°)

= 3 3 1 1- × + -

2 2 2 2

= R.H.S

EXAMPLE 11Prove that

cos 90° + θ sec -θ tan 180° - θ= -1

sec 360° - θ sin 180° + θ cot 90° - θ .

SOLUTIONL.H.S

cos 90° + θ sec -θ tan 180° - θsec 360° - θ sin 180° + θ cot 90° - θ

= -sinθ secθ -tanθsecθ -sinθ tanθ = – 1 = R.H.S

SECTION - FTRIGONOMETRIC RATIOS OF STANDARD ANGLES

0°

0

1

0

Not defined

1

Not defined

30° 45° 60° 90°

1

0

Not defined

Not defined

0

1

12

122

3

3

2

2

13

13

23

12

1

1

2

23 2

3

SECTION - GADDITION & SUBTRACTION FORMULAE

(a) sin (A ± B) = sinA cosB ± cosA sinB(b) cos (A ± B) = cosA cosB sinA sinB

(c) tan (A ± B) = tan A ± tan B1 ± tan A tan B

(d) cot (A ± B) =cot A cot B ± 1cot B ± cot A

Some More Results :(a) sin (A + B).sin (A – B) = sin2 A– sin2 B = cos2 B – cos2 A

(b) cos (A + B).cos (A – B) = cos2 A – sin2 B = cos2 B – sin2 A

(c) sin (A + B + C) = sin A cos B cos C + cosA sin B sin C+ cos A cos B sin C – sin A sin B sin C

(d) cos (A + B + C)= cos A cos B cos C–cos A sin B sin C– sin A cos B sin C – sin A sinB cosC

(e) tan (A + B + C) =tan tan tan tan tan tan

tan tan tan tan tan tanA B C A B C

A B B C C A

1



SOLVED EXAMPLE

EXAMPLE 12Suppose x and y are real numbers such that tan x + tan y =42 and cot x + cot y = 49. Find the value of tan(x + y).

SOLUTIONtan x + tan y = 42 and cot x + cot y = 49

tan(x + y) = tanx + tany1 - tanx tany

now, cot x + cot y = 49

1 1+

tanx tany = 49

tany + tanxtanx · tany = 49

tan x · tan y =tanx + tany

49 = 49

42 =

67

tan (x + y) =42

1- (6 7) =421 7 = 294 Ans.

EXAMPLE 13

If x sin = y sin 2πθ +3

= z sin 4πθ +3

then :

SOLUTION

x sin = y 1 3- sinθ + cosθ2 2

xy

=3

2 cot

12

similarly xz

= 3

2 cot

12

on addingxz

+xy

= 1

xy + yz + zx = 0 Ans. B

EXAMPLE 14If tan A & tan B are the roots of the quadratic equation, ax2 + b x + c = 0 then evaluatea sin2 (A+ B)+b sin (A + B) . cos (A + B) + c cos2 (A + B).

SOLUTION

tan A + tan B = ba

; tan A . tan B = ca

tan (A + B) =b- ac1 - a

= b

c - a

NowE = cos2 (A + B) [a tan2 (A + B) + b tan (A + B) + c]

= 22

1b1 + (c - a)

2 2

2a b b+ + c

c - a(c - a)

=2

2 2(c - a)

b + (c - a)

2b a + 1 + cc - a c - a

=2

2 2(c - a)

b + (c - a)

2

2b c + c

(c - a)

E = c

EXAMPLE 15Show that cos2A + cos2(A + B) + 2 cosA cos(180° + B) ·cos(360° + A + B) is independent of A. Hence find itsvalue when B = 810°.

SOLUTIONcos2A + cos2(A + B) – [2 cosA · cosB · cos (A + B)]cos2A + cos2(A + B) – [ {cos(A + B) + cos(A – B) } cos(A + B) ]cos2A + cos2(A + B) – cos2(A + B) – (cos2A – sin2B)= sin2B which is independent of Anow, sin2(810°) = sin2(720° + 90°) = sin290° = 1 Ans.

EXAMPLE 16Simplify: cos x · sin(y – z) + cos y · sin(z – x) + cos z · sin(x – y) where x, y, z R.

SOLUTION(1/2)[sin(y – z + x) + sin(y – z – x) + sin(z – x + y) +sin(z – x – y) + sin(x – y + z) + sin(x – y – z)] = 0

SECTION - HTRANSFORMATION FORMULAE

(a) sin C + sin D = 2 sin C DF

HGIKJ2 . cos

C DFHG

IKJ2

similarly other formula are,

(b) sin C – sin D = 2 cosC DF

HGIKJ2 . sin

C DFHG

IKJ2

(c) cos C + cos D = 2 cosC DF

HGIKJ2 . cos

C DFHG

IKJ2

(d) cos C – cos D = – 2 sinC DF

HGIKJ2 . sin

C DFHG

IKJ2

FORMULAE FOR PRODUCT INTO SUM ORDIFFERRENCE CONVERSIONFormulae :(a) 2 sin A cos B = sin (A + B) + sin (A – B)(b) 2 cos A sin B = sin (A + B) – sin (A – B)(c) 2 cos A cos B = cos (A + B) + cos (A – B)(d) 2 sin A sin B = cos (A – B) – cos (A + B)



SOLVED EXAMPLE

EXAMPLE 17

Calculate 4 sinπ1+6

cos π1+3

.

SOLUTION

4sinπ1+6

cosπ1+3

= 2 π π π πsin 1+ +1+ + sin 1+ -1-6 3 6 3

= 2 π πsin 2 + + sin -2 6

= 2π 1sin - (-2) -2 2

= 2cos (–2) – 1 = 2 cos 2 – 1.

Thus, 4 sinπ1+6

cosπ1+3

= 2 cos 2 – 1.

EXAMPLE 18Prove that cos55° + cos65° + cos175° = 0

SOLUTIONL.H.S = cos55° + cos65° + cos175°

= 55° + 65° 55° - 65°2cos cos + cos175°

2 2

= 2cos60°cos(–5°) + cos175° = 12× cos5°2 +cos(180°– 5)

= cos5° – cos5° = 0

EXAMPLE 19Prove that cos18° – sin18° = 2sin27° .

SOLUTIONcos18° – sin18°= cos18° – sin(90° – 72°) = cos18° – cos72°

= 18° + 72° 72° -18°2sin sin

2 2

= 2sin45°sin27° = 12 sin27°2 =

2sin27°

SECTION - IMULTIPLE ANGLE / SUBMULTIPLE ANGLE

FORMULAEMultiple angle/Submultiple angle formulae

(i) sin 2 = 2 sin cos =

2tan1

tan2

(ii) cos 2 = cos2 – sin2 = 2 cos2 – 1

= 1 – 2 sin2 =

2

2

tan1tan1

(iii) tan 2=

2tan1

tan2

(iv) sin 3 = 3 sin – 4 sin3

(v) cos 3 = 4 cos3 – 3 cos

(vi) tan 3 =

2

3

tan31tantan3

(vii) sin /2 = 2cos1

(viii) cos /2 = 2cos1

(ix) tan /2 =

cos1cos1

=

sincos1

=

cos1

sin

SOLVED EXAMPLE

EXAMPLE 20If cot = 1/2, then find the values of sin2 and cos2.SOLUTION

sin 2 = 22tanθ

1+ tan θ =

2·21+ 4

= 54

;

cos 2 =2

21- tan θ1+ tan θ

=1- 41+ 4 = –

35

EXAMPLE 21

Prove thattan8θtanθ

= (1 + sec2) (1 + sec4) (1 + sec8).

SOLUTION

RHS =1+ cos2θ 1+ cos4θ 1+ cos8θ× ×

cos2θ cos4θ cos8θ

= 2 2 22cos θ × 2cos 2θ × 2cos 4θcos2θ cos4θ cos8θ

=[8cosθcos2θcos4θ]cosθ

cos8θ =

sin8θ cosθsinθ

cos8θ

EXAMPLE 22Prove the identity,

cos3π + 4α2

+ sin (3 8) sin (412)

= 4 cos 2 cos 4 sin 6 .



SOLUTIONLHS : sin 4 + sin 8 + sin 12= 2 sin 8 cos 4 + sin 8 = 2 sin 8 cos 4 + 2sin4 cos 4= 2 cos 4[sin 8 + sin 4] = 2 cos 4 [2 sin 6 cos2]= 4cos 2 cos 4 sin 6

EXAMPLE 23

If cos =2 cosβ - 12 - cosβ then find the value of tan

2

cotβ2

(0

< < and 0 < < )

SOLUTION

1cos α =

2 - cosβ2 cosβ - 1

1 - cos α1 + cos α

= 3 1 - cosβ1 + cosβ

(Componendo & dividendo)

tan2α2

= 3 tan2β2

tan2α2

cot2β2

= 3 Ans. 3

EXAMPLE 24

Calculate cosα2

if sin =45

and 3π- , -π2

.

SOLUTIONFirst of all we seek cos . Since cos is negative for anyangle belonging to the indicated interval, we have cos

= – 21 - sin α = – 35

.

Since 3π- , -π2

, it follows that α 3π πÎ - , -2 4 2

.

For any angleα2

belonging to this interval cosα2

is also

negative, and therefore cosα2

= –1 + cosα 5= -

2 5 .

Thus cos α 5= -2 5

.

EXAMPLE 25

Calculate tanα2

if cos 2 =7

32 and

3π-π, -4

.

SOLUTIONSince cos is negative for an angle belonging to theindicated interval,

we have cos =1 + cos2α

2 = –398

.

Since 3π-π, -4

, it follows thatα2

π 3π- , -2 8

.

For any angleα2

belonging to this interval tan2

is negative,

and therefore α 1 - cosαtan -2 1 + cosα

tanα2

= – 8 + 39

5.

EXAMPLE 26

If tanθ

tanθ - tan3θ = 31

, find the value of cotθ

cotθ - cot3θ .

SOLUTION

tanθtanθ - tan3θ

= 31

3 tan = tan – tan 3

Þ 2 tan + tan 3 = 0 2 tan +3

23tanθ - tan θ

1- 3tan θ = 0,

2(1 – 3 tan2) + 3 – tan2 = 0 tan2 = 57

now,

cotθcotθ - cot3θ =

tan3θtan3θ - tanθ

=

3

32

2

3tanθ - tan θ3tanθ - tan θ(1 - 3tan θ) - tanθ

1 - 3tan θ

=2 2

2 2 2tanθ(3 - tan θ)(1 - 3tan θ)

tanθ(1 - 3tan θ)(3 - tan θ -1 + 3tan θ)

=2

23 - tan θ

2(1+ tan θ) =

3- (5 7)2 1+ (5 7) =

162·12 =

23 Ans.Ans.

Alternatively:

Prove thattanθ

tanθ - tan3θ +

cotθcotθ - cot3θ

= 1

now proceed



EXAMPLE 27Prove using induction or otherwise that,

2 cos2n

= 2 + 2 + 2 + ...... 2 + 2cos θ

where R. H. S. contains n radical signs and (0 , ).

SOLUTION

2 cos2

= 2 (1 + cosθ)

2 cos 2θ2

= θ22 1+ cos = 2 + 2 (1 + cosθ)

2 cos 3θ2

= 2θ22 1 + cos= 2

θ22 + 2cos = 2 + 2 + 2 (1 + cosθ) and so on.

In the same way 2 cosnθ2

=

2 + 2 + 2 + ...... + 2 + 2 cosθ

Similarly 2 sinnθ2

= 2n -1θ

21 - cos

2 = n-1

θ2 - 2cos2

= 2 - 2 + 2 + 2 + ...... 2 + 2 cosθ

where R. H. S. contains n radical signs

SECTION - JMORE STANDARD ANGLES

(i) sin 15° or sin π12

=3 - 1

2 2 = cos 75° or cos

5π12

cos 15° or cos π12

=3 + 12 2

= sin 75° or sin 125

tan 15° =3 - 13 + 1

= 2 3 = cot 75° ;

tan 75° = 3 + 13 - 1

= 2+ 3 = cot 15°

(ii) sinπ8 =

2 - 22

; cosπ8 =

2 + 22

;

tanπ8 =

2 - 1 ; tan3π8 =

2 + 1

(iii) sinπ

10 or sin 18° =5 - 14

& cos 36°

or cosπ5

= 5 + 14

SOLVED EXAMPLE

EXAMPLE 28

If x = 7.5° then find the value of cosx - cos3xsin3x - sinx

.

SOLUTIONcosx - cos3xsin3x - sinx =

2sin2xsinx2sinxcos2x = tan 2x = tan (2 ×7.5)

= tan 15° = 2 – 3 Ans.

EXAMPLE 29

Prove that cos36°cos72°cos108°cos144° = 1

16 .

SOLUTIONcos36°cos72°cos108°cos144°= cos36°sin18°(–sin18°)(–cos36°)

= cos236°sin218° = 2 2

5 +1 5 -14 4

= 2

5 + 1 5 -14 4

= 1

16

SECTION - KTRIGONOMETRIC SERIES

(i) An Increasing Product series :p = cos. cos 2 . cos 22 ....... cos (2n–1

)

=

π1) (2k αif1,2kπαif1,

nπαif,sinα2

αsin2n

n

(ii) sin sin (60º – ) sin (60º + ) = 41

sin 3

(iii)) cos cos (60º – ) cos (60º + )=41

cos 3(iv) tan 3 = tan.tan(60º – ).tan (60º + )

SOLVED EXAMPLE

EXAMPLE 30

Prove that sin6°sin42°sin66°sin78° = 1

16 .

SOLUTIONsin6°sin42°sin66°sin78°= sin6°cos48°cos24°cos12°

= sin6°3

3

2 sin12°cos12°cos24°cos48°2 sin12°

= 3sin96°sin6°

2 sin12°

= 42sin6°cos6°

2 sin12° = 4sin12°

2 sin12° = 1

16



EXAMPLE 31

Prove that sin10°sin30°sin50°sin70° = 1

16SOLUTIONsin10°sin30°sin50°sin70°=sin10°sin(60°- 10°)sin(60° + 10°)sin30°

= 1 sin 3×10° sin30°4 = 21 Sin 30°

4 = 1

16

SECTION - LGRAPHS OF TRIGONOMETRIC FUNCTIONS

sin=X–2

–3/2 –

–/2 /2 3/2O 2 X

y = 1

y = –1

Y

Y

cos=X –3/2 – –/2 /2 3/2O X

y = 1

y = –1

Y

Y

tan=X X3/2

/2

O–/2

––3/2

Y

Y

cot=X–2

–3/2–

–/2 /2

3/2

O 2 X

Y

Y

sec=X

(–2,1) (2,1)–3/2–3/2

–2 ––/2

O

(0,1)

/2 3/2 5/22

(–1)(––1)

Y

Y

X

y = –1

y = 1

cosec=X

y = 1

y = –1

Y

(–3/2,1) (/2, 1)

O(–/2,–1) (3/2,–1)

Y

X

SECTION - MRANGE OF TRIGONOMETRIC FUNCTIONS

(a) Min. value of a2 tan2 + b2 cot2 = 2ab

(b) Max and Min. value of acos + bsin are 2 2a + b

and – 2 2a + b(c) If f() = acos() + bcos() where a, b, and

are known quantities then

– 2 2a + b + 2abcos(α - β) f()

2 2a + b + 2abcos(α - β)

(d) If π0,2

and = (constant) then the

maximum values of the expressioncos cos, cos + cos, sin + sin and sin sinoccurs when /2

(e) If π0,2

and = (constant) then the

minimum values of the expressionsec + sec, tan + tan, cosec + cosec occurswhen /2.

(f) If A, B, C are the anlges of a triangle then maximumvalue of sinA + sinB + sinC and sinA sinB sinCoccurs when A = B = C = 60º

(g) In case a quadratic in sin or cos is given then themaximum or minimum values can be interpretedby making a perfect square



SOLVED EXAMPLE

EXAMPLE 32Find the minimum vertical distance between the graphs ofy = 2 + sin x and y = cos x.SOLUTION

dmin = min(2 + sin x – cos x) = min[2 + 2 sinπx -4

]

= 2 – 2 at x = 47

EXAMPLE 33Find the minimum and maximum value off (x, y) = 7x2 + 4xy + 3y2 subjected to x2 + y2 = 1.

SOLUTIONLet x = cos and y = sin y = f () = 7 cos2 + 4 sin cos + 3 sin2= 3 + 2 sin 2 + 2(1 + cos 2)= 5 + 2(sin 2 + cos 2)but – 2 (sin 2 + cos 2) 2

ymax = 5 + 22 and ymin = 5 – 22

EXAMPLE 34If 1, 2, ...... , n are real numbers, show that,(cos 1 + cos 2 +... + cos n)

2 + (sin 1 + ...+ sin n)2 n2

SOLUTIONL H S = (cos

2 1 + sin2 1) + ....... + (cos

2 n + sin2 n) +

2 cos (1 2)

nC2 terms n + 2n (n - 1)

2 = n2

EXAMPLE 35

Show that the expression cos (sin + 2 2sin θ + sin α )

always lies between the values of ± 21 + sin α .SOLUTION

Let y = cos (sin + 2 2sin θ + sin α )

or, y – cos sin = cos ( 2 2sin θ + sin α )or, (y – cos sin)2 = cos2 (sin2 + sin2)or, y2 – 2ysin cos + cos2 = cos2 sin2 + cos2 sin2or, y2 – 2ysin cos + cos2 = cos2 + cos2 . sin2[Here we have added cos2 on both sides to get 1 + sin2]or, y2 – 2y sin cos + cos2 = cos2 (1 + sin2)or, y2.sec2 – 2y tan + 1 = 1 + sin2 (dividing by cos2 )

or, y2tan2 – 2ytan + 1=(1+sin2) – y2(sec2 = 1 + tan2 )or, (ytan – 1)2 = (1 + sin2) – y2 square of a real number 0 1 + sin2 – y2 0

or, y2 – ( 21 + sin α )2 0

y lies between – 21 + sin α and 21 + sin α .

SECTION - NSUMMATION SERIES PROBLEMS

Sum of sines or cosines of n anglessin + sin (+) + sin ( + 2 ) + ...... + sin (α + n -1 β)

= nβ2β2

sin

sin sin

n -1α + β2

cos + cos (+) + cos ( + 2 ) + ...... + cos (α + n -1 β)

= nβ2β2

sin

sin cos

n -1α + β2

SOLVED EXAMPLE

EXAMPLE 36Find the sum of the series ,cot 2 x . cot 3 x + cot 3 x . cot 4 x + ...... + cot (n + 1) x .cot (n + 2) x .

SOLUTIONcot x = cot [ (n + 2) x (n + 1) x ]

= cot (n + 2) x . cot (n + 1) x + 1

cot (n + 1) x - cot (n + 2) x

or cot x [ cot (n + 1) x cot (n + 2) x ] = cot (n + 2) x . cot(n + 1) x + 1Hence cot (n + 1) x cot (n + 2) x = cot x [ cot (n + 1) x cot (n + 2) x ] 1Put n = 1 , 2 , 3 , ...... , n and adding we get sum of the series

= cot x [ cot 2 x cot (n + 2) x ] n2

EXAMPLE 37Let f (x) denote the sum of the infinite trigonometric Series,

f (x) = ¥

n nn=1

2x xsin sin3 3 .

Find f (x) (independent of n) also evaluate the sum of thesolutions of the equation f (x)=0 lying in the interval(0, 629).



SOLUTION

f (x) = n nn=1

2x xsin sin3 3

= n nn=1

1 2x x2sin sin2 3 3

= ¥

n n-1n=1

1 x xcos - cos2 3 3

now substituting n = 1, 2, 3, 4........

f(x)=1 xcos - cosx2 3

+ 21 x xcos - cos2 33

+

3 21 x xcos - cos2 3 3

.......... + n n-11 x xcos - cos2 3 3

f (x) = nn1 xLim cos - cosx2 3

= 21

[1 – cos x] now

f (x) = 0 cos x = 1 x = 2n, n Isum of the solutions in (0, 629),S = 2[ + 2 + 3 + ....... + 100] = 2 · 5050 = 10100

EXAMPLE 38

Evaluate 89

2n=1

11+ (tann°) .

SOLUTION

S = 21

1+ (tan1°) + 21

1+ (tan2°) + 21

1+ (tan3°) + ....+

2

11+ (tan88°) + 2

11+ (tan89°)

reversing the sum

S = 21

1+ (cot1°) + 21

1+ (cot2°) + ..... + 21

1+ (cot88°)

+ 21

1+ (cot89°)

2S = 89

2 2n=1

1 1+1+ (tann°) 1+ (cotn°)

= 289

2 2n=1

1 (tann°)+1+ (tann°) 1+ (tann°)

=89

n=11 = 1 + 1 + ....... + 1 = 89 S = 44.5

SECTION – OMIXED PROBLEMS

CONDITIONAL IDENTITIEStan (A+B+C) =

tan A + tan B + tan C - tan A tan B tan C1 - tan A tan B - tan B tan C - tan C tan A

If A+B+C = then(a) tanA + tanB + tanC = tanA tanB tanC

(b) tanA2

tanB2

+ tanB2

tanC2

+ tanC2

tanA2

= 1

(c) sin2A + sin2B + sin2C = 4 sinA sinB sinC

(d) sinA + sinB + sinC = 4 cosA2 cos

B2 cos

C2

(e) cos 2A + cos 2B + cos 2C = –1 – 4 cos A cos B cos C

(f) cos A + cos B + cos C = 1 + 4 sinA2

sinB2

sin C2

SOLVED EXAMPLE

EXAMPLE 39Eliminate between the equation a sec + b tan + c = 0and p sec + q tan + r = 0.

SOLUTIONGiven a sec + b tan+ c = 0 ...(1)

and p sec + q tan+ r = 0 ...(2)

Solving (1) and (2) by cross multiplication method, we have

secθ tanθ 1= =br - qc pc - ar aq - pb sec

2 – tan2 = 1

2 2

br - qc pc - ar-aq - pb aq - pb

= 1

or, (br – qc)2 – (pc – ar)2 = (aq – pb)2

EXAMPLE 40If is eliminated from the equations, a cos + b sin = c& a cos2 + b sin2 = c, show that the eliminant is, (a b)2(a c) (b c) + 4 a2 b2 = 0 .

SOLUTIONa cos + b sin = c ...(1)a cos2 + b sin2 = c ...(2)

From (2) sin2 =c - ab - a and cos

2 = b - cb - a

Now squaring (1) a2 cos2 + b2 sin2 + 2 ab sin cos = c2



a2b - cb - a + b

2c - ab - a c

2 = 2 ab b - c c - ab - a b - a

ora2 (b c) + b2 (c a) c2 (b a)

= 2 ab b - c c - a (a b) (b c) (c a)

= 2 ab b - c c - a

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

= 4 a2 b2 (b c) (c a)(a b)2 (b c) (c a) = 4 a2 b2 Result

EXAMPLE 41

Prove that tanπ7

. tan 2p7

. tan3π7

= 7

SOLUTION

Let =π7

7 =

or, 4 + 3 = or, tan(4) = tan( – 3)or, tan4 = –tan3

or,3

2 44tanθ - 4tan θ

1 - 6tan θ + tan θ = –

3

23tanθ - tan θ

1 - 3tan θ

or,3 3

2 4 24z - 4z 3z - z= -

1- 6z + z 1- 3z

[where tan = z (suppose)]or, (4 – 4z2) (1 – 3z2) = –(3 – z2)(1 – 6z2 + z4)or 12z4 – 16z2 + 4 = –(–z6 + 9z4 – 19z2 + 3)or, z6 – 21z4 + 35z2 – 7 = 0 ...(1)

This is cubic equation in z2 i.e. in tan2, the roots of this

equation are therefore tan2π7

, tan22π7

and tan23π7

From (1), product of the roots = 7

tan2 π7

. tan22π7

. tan23π7

= 7

tan π7

. tan2π7

. tan3π7

= 7 Hence the result.

EXAMPLE 42In triangle ABC, cos A . cos B + cos B . cos C + cos C .cos A = 1 – 2 cos A . cos B . cos C. Prove that it is possibleif and only if ABC is equilateral.

SOLUTION cos A . cos B = 1 – 2 cos A . cos B . cos C= 1 – cos C (cos (A+ B) + cos (A – B) )= 1 – cos C (cos (A – B) – cosC)= 1 + cos (A + B) cos (A – B) +cos2 C= 1 + cos2 A – sin2 B + cos2C= cos2 A + cos2 B +cos2C = cos2A.Thus we have, 2 cos2 A – 2 cos A . cos B = 0 (cos A– cos B)2 + (cos B–cos C)2+(cos C– cos A)2=0 cos A = cos B = cos C A = B = CThus triangle ABC is equilateral

Now if is equilateral A = B = C =π3

cosA cos B =34

and 1 – 2 cos A cos B cos C

= 1 – 2 3=8 4

. Hence the given expression is true if and

only if ABC is equilateral.

Trignometric Ratios & Identities 4.13


SECTION - AANGLE AND ITS UNITS

1. Which is correct one ?(A) sin 1° < sin 1 (B) sin 1° = sin 1

(C) sin 1° > sin 1 (D) sin 1° = sin

180

2. The difference between two acute angles of a right

angle triangle is 3

10 rad. Find the angles in degree.

(A) 54°, 36° (B) 75°,15°(C) 72°, 18° (D) None of these

SECTION - BBASIC DEFINATION OF TRIGONOMETRIC RATIO

3. A regular hexagon & a regular dodecagon areinscribed in the same circle. If the side of the

dodecagon is ( 3 –1), then the side of the hexagonis

(A) 2 +1 (B) 3 +12

(C) 2 (D) 2

4. Calculate tan if cos = –55

and

23, .

(A) –2 (B) 2(C) 1 (D) None of these

SECTION - CSIGNS OF TRIGONOMETRIC RATIOS

5. The sign of the product sin 2 sin 3 sin 5 is -(A) Negative (B) Positive(C) 0 (D) None of these

6. If sin = 1213

, 2 , find the value of sec +

tan(A) 5 (B) 3(C) – 3 (D) –5

SECTION - DFUNDAMENTAL IDENTITIES

7. If sin6 + cos6 – 1 = sin2cos2, then the valueof is(A) 1 (B) 2(C) 3 (D) None of these

8. If a cos + b sin = 3 & a sin – b cos = 4 thena2 + b2 has the value =(A) 25 (B) 14(C) 7 (D) None of these

9. If 4 4sin α cos α 1+ =

a b a + b , find 38

3

8

bcos

asin

(A) 31

(a b) (B) 31

(a b)

(C) 3 31 1a b

(D) None of these

SECTION - EREDUCTION FORMULAE

10.3π 3π 7πtan x - cos + x - sin - x

2 2 2π 3πcos x - .tan + x2 2

when simplified reduces to :(A) sinx cosx (B) – sin2x(C) –sinx cosx (D) sin2x

11. The expression

3

)3(sin2

3sin 44 –

2

)5(sin2

sin 66 is equal to

(A) 0 (B) 1(C) 3 (D) sin 4 + sin 6

12. The value of sin() sin () cosec2 is equal to(A) –1 (B) 0(C) sin (D) None of these

SECTION - FTRIGONOMETRIC RATIOS OF STANDARD ANGLES

13. The value of tan 1º tan 2º tan 3º ..... tan 89º is(A) 1 (B) 0(C) (D) 1/2

14. The value of the expression cos 1° cos 2° .........cos 179° equals(A) 0 (B) 1(C) 1/ 2 (D) – 1

Exercise - 1 Objective Problems | JEE Main



SECTION - GADDITION & SUBSTRACTION FORMULAE

15. If sin sin – cos cos + 1 = 0, then the value of1 + cot tan is(A) 1 (B) –1(C) 2 (D) None of these

16. The value of cot x + cot(60º + x) + cot (120º + x) isequal to(A) cos 3x (B) tan 3x

(C) 3 tan 3x (D) 2

3

3 - 9 tan x3tanx - tan x

17. If tan A – tan B = x and cot B – cot A = y, then cot(A – B) is equal to

(A) 1 1-y x (B)

1 1-x y

(C) y1

x1 (D) None of these

18. If tan 25º=x, thentan155º -tan115º

1 + tan155º tan115º is equal to

(A) 21 - x

2x(B)

21+ x2x

(C) 2

2

1+ x1- x

(D) 2

2

1- x1+ x

19. Let be such that 3. If sin + sin = – 2165

and cos + cos = –2765 , then the value of

cos α - β

2 is

(A) – 3130

(B) 3

130

(C) 6

65 (D) – 6

65

20. The value of tan 3A – tan 2A – tan A is equal to(A) tan 3A tan 2A tan A(B) – tan 3A tan 2A tan A(C) tan A tan 2A – tan 2A tan 3A – tan 3A tan A(D) None of these

21. If = 2, then

(A) tan 2

+tan2

+ tan2

= tan 2

tan 2

tan 2

(B) tan 2

tan 2

+ tan 2

tan 2

+ tan 2

tan 2

= 1

(C) tan 2

+ tan 2

+ tan 2

= – tan 2

tan 2

tan 2

(D) tan 2

tan 2

+ tan 2

tan 2

+ tan 2

tan 2

= 0

22. In a triangle ABC if tan A < 0 then :(A) tan B . tan C > 1 (B) tan B . tan C < 1(C) tan B . tan C = 1 (D) None of these

23. If A + B = 225º, then the value of

cotA cotB.1+ cotA 1+ cotB

is

(A) 2 (B) 1/2(C) 3 (D) 1/3

24. The value of sin24 º co s6º -sin 6º sin6 6º

sin 21 º cos3 9º -cos51 º sin 69 º is

(A) –1 (B) 1(C) 2 (D) None of these

SECTION - HTRANFORMATION FORMULAE

25. The numerical value of sin 12º . sin 48º . sin 54º isequal to(A) 1/2 (B) 1/4(C) 1/16 (D) 1/8

26. If A + B + C = 3π2

, then cos 2A + cos 2B + cos 2C

is equal to(A) 1–4cos A cosB cosC(B) 4 sinA sinB sinC(C) 1+2 cosA cosB cosC(D) 1–4 sinA sinB sinC

27. The expression

cos6 x + 6cos4x + 15 cos2 x + 1 0cos5x + 5cos3x + 10 cosx is equal to

(A) cos 2x (B) 2 cos x(C) cos2 x (D) 1 + cos x



SECTION - IMULTIPLE ANGLE / SUBMULTIPLE ANGLE

FORMULAE28. The value of cos 10° – sin 10° is

(A) Positive (B) Negative(C) 0 (D) 1

29. If 0 < x < and cos x + sin x = 21

, then tan x is

(A) (4 - 7 )

3(B) –

(4 + 7 )3

(C) (1+ 7)

4(D)

(1 - 7 )4

30. If A lies in the third quadrant and 3 tan A – 4 = 0,then 5 sin 2A + 3 sinA + 4 cosA is equal to

(A) 0 (B) –524

(C) 524

(D) 548

31. For –π2



42. Find the greatest value of c such that system ofequations x2 + y2 = 25; x + y = chas a real solution(A) 5 (B) 5 2(C) 2 2 (D) 2

SECTION - NSUMMATION SERIES PROBLEMS

43. Given35

k=1sin5k = tan mn

, where angles are

measured in degrees, and m and n are relatively

prime positive integers that satisfynm

< 90, find the

value of (m + n).(A) 125 (B) 2(C) 177 (D) 179

44. The value of cos 11

+ cos11

+ cos5

11 + cos

711

+

cos 911

is

(A) 0 (B) 1

(C) 12

(D) None of these

SECTION – OMIXED PROBLEMS

45. If tan +cot =a then the value of tan4+cot4 =(A) a4 + 4a2 +2 (B) a4 – 4a2 + 2(C) a4 – 4a2 – 2 (D) None of these

46. The number of real solutions of the equationsin (ex) = 2x + 2–x is -(A) 1 (B) 0(C) 2 (D) Infinite

47. In a right angled triangle the hypotenuse is 2 2times the perpendicular drawn from the oppositevertex. Then the other acute angles of the triangleare

(A) &3 6

(B) 3&

8 8

(C) &4 4

(D) 3&

5 10



ANGLE AND ITS UNITS/BASIC DEFINATION OF TRIGONOMETRIC RATIO/SIGNS OF TRIGONOMETRIC RATIOS/FUNDAMENTAL IDENTITIES

1. Find the length of an arc of a circle of radius 6 cmsubtending an angle of 15° at the centre.

(A) cm4

(B) cm

(C) cm2

(D) None of these

2. If sin + cosec = 2, then the value of sin20 +cosec20 is(A) 1 (B) 2(C) 3 (D) 4

3. If sin = –2

1 and tan = 1, then lies in which

quadrant-(A) First (B) Second(C) Third (D) Fourth

4. tan 7 cos 11 sin 27 is(A) negative(B) positive(C) neither negative nor positive(D) None of these

5. If cot + tan = m and cos

1 – cos = n, then

(A) m (mn2)1/3 – n(nm2)1/3 = 1(B) m(m2n)1/3 – n(nm2)1/3 = 1(C) n (mn2)1/3 – m(nm2)1/3 = 1(D) n(m2n)1/3 – m(mn2)1/3 = 1

6. If 4

3 < , then 2

12cotsin

is equal to

(A) 1 +cot (B) –1 – cot (C) 1 – cot (D) –1 + cot

7. If 2 sec2 – sec4 – 2 cosec2 + cosec4 = 15/4,then tan is equal to(A) 2/1 (B) 1/2

(C) 1/2 2 (D) 1/4

8. If 3 sin x + 4 cos x = 5 then 4sin x – 3 cos x is equal to(A) 0 (B) 1(C) 5 (D) None of these

REDUCTION FORMULAE/TRIGONOMETRIC RATIOS OF STANDARD ANGLES/ADDITION & SUBSTRACTION FORMULAE/ADDITION & SUBSTRACTION FORMULAE/TRANFORMATION FORMULAE/

9. cos0+cos 7 +cos 7

2 +cos 73 +cos 7

4 +cos 75 +cos 7

6 =

(A) 1/2 (B) –1/2(C) 0 (D) 1

10. If x sin45° cos260° = tan cosecsec cot

2

2

60 3045 30

, then x

is equal to(A) 9 (B) 10(C) 11 (D) None of these

11. If A + B + C = & cos A = cos B . cos C thentan B . tanC has the value equal to(A) 1 (B) 1/2(C) 2 (D) 3

12. If 3 sin = 5 sin, then

α + βtan2

α - βtan2

is equal to

(A) 1 (B) 2(C) 3 (D) 4

MULTIPLE ANGLE / SUBMULTIPLE ANGLEFORMULAE

13. If f() = sin2 + sin2 2πθ +3

+ sin2 4πθ +3

,

then f

15 is equal to

(A) 32

(B) 23

(C) 31

(D) 21

14. If cosec A + cot A = 211

, then tan A is

(A) 2122

(B) 1516

(C) 44

117(D)

11743

Exercise - 2 (Level-I) Objective Problems | JEE Main



15. If sin 2 = k, then the value of3 3

2 2

tan θ cot θ+1+ tan θ 1 + cot θ

is equal to

(A) 21 - k

k(B)

22 - kk

(C) k2 + 1 (D) 2 – k2

16. If cos + cos = a, sin + sin = b and = 2,

then cos3θcosθ

=

(A) a2 + b2 – 2 (B) a2 + b2 – 3(C) 3 – a2 – b2 (D) (a2 + b2) /4

17. The value of 2

2

1- tan 15º1 + tan 15º

is

(A) 1 (B) 3

(C) 3

2(D) 2

18. If tan2 = 2 tan2 + 1, then the value of cos 2 +sin2 is(A) 1 (B) 2(C) –1 (D) Independent of

19. If π , π2

then the value of

1+ sinα – 1 - sinα is equal to

(A) 2 cos α2

(B) 2 sin α2

(C) 2 (D) None of these

20. 4cos222 ; 0 < < /4 is(A) cos (B) sin (C) 2 cos (D) 2 sin

MORE STANDARD ANGLES/TRIGONOMETRICSERIES/GRAPHS OF TRIGONOMETRIC FUNCTIONS/RANGE OF TRIGONOMETRIC FUNCTIONS/SUMMA-TION SERIES PROBLEMS

21. The value of tan π

16 + 2 tan

π8

+ 4 is equal to

(A) cotπ8

(B) cotπ

16

(C) cotπ

16–4 (D) None of these

22. If A = tan 6º tan 42º and B = cot 66º cot 78º, then(A) A = 2B (B) A = 1/3 B(C) A = B (D) 3A = 2B

23. The value of

6cos

12cos

24cos

48cos

48sin364 is

(A) 8 (B) 6 (C) 4 (D) 12

24. Let a and b be positive numbers not equal

to 1 and 22

3.

If atan > btan > 1, then which one of thefollowing is always TRUE ?(A) a > b > 1 (B) a < b < 1(C) b < a < 1 (D) b > a > 1

25. If f() = sin4 + cos2 , then range of f() is

(A) 1 ,12

(B) 1 3,2 4

(C) 3 ,14

(D) None of these

26. The value of cos19

+cos3π19

+cos5π19

+...+

cos17π19

is equal to(A) 1/2 (B) 0(C) 1 (D) None of these

27. Find the sum of the series, cosπ

2n +1 + cos

3π2n +1

+ cos5π

2n +1 + ........ upto n terms.

Do not use any direct formula of summation.

(A) 0 (B) 12

(C) –12 (D) None of these

MIXED PROBLEMS

28.1 1+

cos290º 3sin250º =

(A) 2 3

3(B)

4 33

(C) 3 (D) None of these



29. In any triangle ABC, which is not right angled cos A . cosec B . cosec C is equal to(A) 1 (B) 2(C) 3 (D) None of these

30. If 3 cos x + 2 cos 3x = cos y, 3 sin x + 2 sin 3x = siny, then the value of cos 2x is(A) –1 (B) 1/8(C) –1/8 (D) 7/8

31. If 0° < x < 90° & cos x = 310 , then the value of

log10 sin x + log10 cos x + log10 tan x is(A) 0 (B) 1(C) –1 (D) None of these

32. If tan =1

12 122 +

where (0, 2), find

the number possible values of .(A) 2 (B) 3(C) 4 (D) more than 4

33. If sinA 3 cosA 5= and =sinB 2 cosB 2

, 0 < A, B



ANGLE AND ITS UNITS/FUNDAMENTAL IDENTITIES/ADDITION & SUBSTRACTION FORMULAE

1. Which of the following is not correct?(A) cos1º > cos1 (B) sin1º = cos1(C) cos1º < cos1 (D) sin1º > sin1

2. The value of 3sinx + cosx

cos x =

(A) 1+tanx + tan2x –tan3x(B) 1+tan x+tan2x+tan3x(C) 1–tanx + tan2x +tan3x(D) (1 + tan x) sec2x

3. If (sec A + tan A) (sec B + tan B) (sec C + tan C)= (sec A – tan A) (sec B – tan B) (sec C – tan C)then each side is equal to(A) 1 (B) –1(C) 0 (D) None of these

4. In a triangle tan A + tan B + tan C = 6 and tan Atan B = 2, then the values of tanA, tan B and tan Care(A) 1, 2, 3 (B) 2, 1, 3(C) 1, 2, 0 (D) None of these

5. For 0 < < /2, tan + tan 2 + tan 3 = 0 if(A) tan = 0 (B) tan 2 = 0(C) tan 3 = 0 (D) tan tan 2 = 2

6. If x + y = z, then cos2 x + cos2 y + cos2 z – 2 cos xcos y cos z is equal to(A) cos2 z (B) sin2 z(C) cos (x + y – z) (D) 1

TRANFORMATION FORMULAE/TRIGONOMETRIC SERIES/RANGE OF TRIGONOMETRIC FUNCTIONS/MIXED PROBLEMS

7. If 3 sin =sin (2+), then tan () – 2 tan is

(A) independent of

(B) independent of

(C) dependent of both and

(D) independent of but dependent of

8. If the sides of a right angled triangle are{cos2 + cos2 + 2cos()} and{sin2 + sin2 + 2sin()},then the length of thehypotneuse is(A) 2 [1 + cos()] (B) 2 [1 – cos()]

(C) 4 cos2 α - β

2(D) 4 sin2

α + β2

9. The value of cosπ

10cos

2π10

cos4π10

cos8π10

cos16π10

is

(A) 10 + 2 564

(B) –cos(π/10)

16

(C) cos(π/10)

16(D) – 10 + 2 5

64

10. An extreme value of 1 + 4 sin + 3 cos is(A) – 3 (B) – 4(C) 5 (D) 6

11. The equation sin6x + cos6x = a2 has real solution if

(A) a (–1,1) (B) a (–1, –1/2)

(C) a 1 1-2 2

(D) a (1/2, 1)

12. If tan x = 2b

a - c, (a c)

y = a cos2x + 2b sin x cos x + c sin2xz = a sin2x – 2b sin x cos x + c cos2x, then(A) y = z (B) y + z = a + c(C) y – z = a – c (D) y – z = (a – c)2 + 4b2

Exercise - 2 (Level-II) Multiple Correct | JEE Advanced



1. Express 1·2 radians in degree measure.

2. Calculate sin if cos= –9

11 and

3ππ,2

.

3. Prove that 3(sin x – cos x)4 + 4(sin6x + cos6x) +6(sin x + cos x)2 = 13

4. Eliminate from the relations a sec = 1 – b tan ,a2 sec2 = 5 + b2 tan2

5. If ax by+

cosθ sinθ = a2 – b2, 2 2

axsinθ bycosθ-cos θ sin θ

=

0. Show that (ax)2/3 + (by)2/3 = (a2 – b2)2/3

6. If Pn = cosn + sinn and Qn = cosn – sinn, thenshow that Pn – Pn – 2 = – sin2 cos2 Pn – 4Qn – Qn–2 = –sin2 cos2 Qn – 4 andhence show thatP4 = 1 – 2 sin2 cos2Q4 = cos2 – sin2

7. If tan = –5/12, is not in the second quadrant,then show thatsin(360º -θ) + tan(90º +θ) 181=-sec(270º +θ) + cosec(-θ) 338

8. If A+B=45º, prove that (1+tan A) (1+tan B) = 2

and hence deduce that tan 221º = 2 -12

9. If sin x + sin y = a & cos x + cos y = b, show that,

sin (x + y) = 2 22ab

a + b and tan

x - y2

= ±2 2

2 24 - a - ba + b

.

10. If x + y + z = π2

show that, sin 2x + sin 2y + sin 2z

= 4 cosx cosy cosz.

11. If x + y = + z, then prove that sin2x + sin2y – sin2z= 2 sin x sin y cos z.

12. If A + B + C = 2S then prove thatcos (S – A) + cos(S – B) + cos (S – C) + cos S =

4cos A2

cos B2

cos C2

13. If A + B + C = 0º then prove that sin 2A + sin 2B +sin 2C = –4 sin A sin B sin C.

14. If tan x = 34

, < x < 3π2

, find the value of

sin x2

and cos x2

.

15. Prove that

2

2

α - π1- cotα4 + cos cot4α

α - π 21 + cot4

sec 9α2

= cosec 4.

16. Prove that 4(cos3 20º+cos340º)=3(cos20º+cos 40º)

17. If tan2 +2 tan. tan 2=tan2+2 tan . tan2, thenprove that each side is equal to 1 or tan = ± tan .

18. Prove that

(i) sec8A -1 tan8A=sec4A -1 tan2A

(ii) cosA + sinAcosA - sinA

– cosA - sinAcosA + sinA

= 2 tan 2A

19. Prove that, sin3x . sin3 x + cos 3 x . cos3 x = cos3 2x.

20. If tan = tanα + tanγ

1 + tanα.tanγ ,

prove that sin 2 = sin2α + sin2γ

1 + sin2α.sin2γ .

21. Prove thatsin2+sin22+sin23+....+sin2n=n2

–sinnθcos(n +1)θ

2sinθ

22. Find the extreme values of

cos x cos 2π + x3

cos 2π - x3

23. Find the maximum and minimum values of(i) cos 2x + cos2 x

(ii) cos2 π + x4

(sin x – cos x)2

24. An maximum value of 1 + 4 sin + 3 cos is(A) – 3 (B) – 4 (C) 5 (D) 6

Exercise - 3 | Subjective | JEE Advanced



25. If is the exterior angle of a regular polygon of nsides and is any constant, then prove thatsin + sin () + ....... up to n terms = 0

26. If –2

< x < 2

and y = log10(tan x + sec x). Then

the expression E =y -y1 0 - 1 0

2 simplifies to one of

the six trigonometric functions. find thetrigonometric function.

27. If the arcs of the same length in two circles subtendangles 75° and 120° at the centre, find the ratio oftheir radii.

28. Find the length of an arc of a circle of radius10 cm which subtends an angle of 45° at the centre.

29. If cos () + cos () + cos () = -32

,

prove thatcos + cos + cos = 0, sin + sin + sin = 0

30. For all in π0,2

show that cos (sin ) > sin (cos )

31. Calculate the following without using trigonometrictables :(i) tan 9º – tan 27º – tan 63º + tan 81º

(ii) cosec 10º – 3 sec 10º

(iii) 2 2 sin10º sec5º cos40º+ - 2sin35º

2 sin5º

(iv) cot 70º + 4 cos 70º

32. If sin () = a & sin () = b (0 < /2)then find the value of cos2 () – 4 ab cos ()

33. If A + B + C = , Prove thattanB tanC+tanC tanA+tanA tanB=1+secA . sec B .secC.

ComprehensionIf cos + cos = a and sin + sin = b and isarithmetic mean between and , then

sin 2 + cos 2 = 1 + 2 2nb(a b)a b

.

where n is some integer then answer the following questions34. The value of n is

(A) 0 (B) 1(C) 2 (D) – 2

35. If for n obtained in above question, sinnA = x, thensin A sin 2A sin 3A sin 4A is a polynomial in x, ofdegree(A) 5 (B) 6(C) 7 (D) 8

36. If degree of polynoimal obtained in previousquestion is p and (p – 5) + sin x, cos x, tan x are inG.P., then cos9x + cos6x + 3 cos5 x – 1 =(A) –1 (B) 0(C) 1 (D) None of these

ComprehensionLet p be the product of the sines of the angles of triangleABC and q is the product of the cosines of the angles.

37. In this triangle tan A + tan B + tan C is equal to(A) p + q (B) p – q

(C) pq (D) None of these

38. tan A tan B + tan B tan C + tan C tan A is equal to

(A) 1 + q (B) 1 q

q

(C) 1 + p (D) 1 p

p

39. The value of tan3 A + tan3 B + tan3 C is

(A) 3 2

3

p 3pqq

(B) 3

3

qp

(C) 3

3

pq (D)

3

3

p 3pqq

Matrix Match Type40. Column - I Column - II(A) If for some real x, then equation (P) 2

x + 1x

= 2 cos holds

then cos is equal to (Q) 1(B) If sin + cosec = 2,

then sin2008 + cosec2008 is equal to (R) 0(C) Maximum value of sin4 + cos4 is(D) Least value of 2 sin2 + 3 cos2 is (S) –1



1. Let A and B denote the statementsA : cos + cos + cos = 0 [AIEEE-2009]

B : sin + sin + sin = 0

If cos +cos +cos =23

,then :

(A) A is false and B is true(B) both A and B are true(C) both A and B are false(D) A is true and B is false

2. Let cos ( + ) = 45

and let sin ( – ) = 513

,

where 0 π4

. Then tan 2 =[AIEEE-2010]

(A) 2516

(B) 5633

(C) 1912

(D) 207

3. If A = sin2x + cos4x, then for all real x :[AIEEE-2011]

(A) 34

A 1 (B) 1316

A 1

(C) 1 A 2 (D) 34

A 1316

4. In a PQR if 3 sin P + 4 cos Q = 6 and4 sin Q + 3 cos P = 1, then the angle R is equal to :

(A) π4

(B) 3π4

[AIEEE-2012]

(C) 5π6

(D) π6

5. The expression tan A

1 - c o tA +

co tA1 - tan A

can be

written as : [JEE MAIN-2013](A) tanA + cotA (B) secA + cosecA(C) sin A cos A + 1 (D) secA cosecA + 1

6. Let fk(x) = 1k

(sink x + cosk x) where x R and

k 1. Then f4(x) – f6(x) equals :[JEE MAIN-2014]

(A) 16

(B) 13

(C) 14

(D) 1

12

7. If 5(tan2x – cos2x) = 2cos 2x + 9, then thevalue of cos 4x is : [JEE MAIN-2017]

(A) –35 (B)

13

(C) 29 (D)

– 79

Exercise - 4 | Level-I Previous Year | JEE Main

For 2019 & 2020 year questions you can visit

@ www.onlinetestseries.motion.ac.in



1. (a) If 4sin x

2+

4cos x3

= 15

, then [JEE 2009, 4+4]

(A) tan2x = 23

(B) 8sin x

8 +

8cos x27

= 1

125

(C) tan2x = 13

(D) 8sin x

8 +

8cos x27

= 2

125

(b) For 0 < < /2, the solution(s) of

6

m=1

(m-1)πcosec θ+4

cosec mπθ + 4

= 4 2

is (are)(A) /4 (B) /6(C) /12 (D) 5/12

2. The maximum value of the expression

2 21

sin θ + 3sinθcosθ + 5cos θ is [JEE 2010]

3. The positive integer value of n > 3 satisfying the

equation 1 1 1= +

π 2π 3πsin sin sinn n n

is

[JEE 2011]

4. Let θ, [0,2π] be such that [JEE 2012]

2 θ θ2cosθ(1 - sin j) = sin θ tan + co t cos - 12 2

3tan(2π - θ) > 0 and - 1 < sinθ < -2

Then cannot satisfy

(A) π0 < <2

(B) π 4π< <2 3

(C) 4π 3π< <3 2

(D) 3π < < 2π2

5. The number of distinct solutions of equation54

cos22x + cos4x + sin4x + cos6x + sin6x = 2in the interval [0, 2] is [JEE 2015]

6. The value of

13

k 1

1(k 1) ksin sin

4 6 4 6

is equal to [JEE 2016]

(A) 3 3 (B) 2(3 3)

(C) 2( 3 1) (D) 2(2 3)

7. Let and be non-zero real numbers suchthat 2(cos – cos ) + cos cos = 1. thenwhich of the following is/are true ?

[JEE 2017]

(A) tan 3 tan2 2

= 0

(B) tan 3 tan2 2

= 0

(C) 3 tan tan2 2

= 0

(D) 3 tan tan2 2

= 0

8. Let a, b, c be three non-zero real numberssuch that the equation

3 a cosx 2bsinx c, x , ,2 2

has two distinct real roots and with +

= .3

Then, the value of ba

is ________

[JEE ADVANCE 2018]

Exercise - 4 | Level-II Previous Year | JEE Advanced



EXERCISE - I1. A 2. C 3. D 4. B 5. A 6. D 7. D8. A 9. A 10. D 11. B 12. A 13. A 14. A15. D 16. D 17. C 18. A 19. A 20. A 21. A22. B 23. B 24. A 25. D 26. D 27. B 28. A29. B 30. A 31. A 32. A 33. C 34. B 35. B36. A 37. C 38. D 39. C 40 A 41. A 42. B43. C 44. C 45. B 46. B 47. B

EXERCISE - IILevel - I (Single correct Option - type Questions)

1. C 2. B 3. C 4. B 5. A 6. B 7. A8. A 9. D 10. D 11. C 12. D 13. B 14. C15. B 16. B 17. C 18. D 19. A 20. C 21. B22. C 23. B 24. B 25. C 26. A 27. B 28. B29. B 30. A 31. C 32. C 33. D 34. A 35. A36. A

Level - II (Multiple correct Option - type Questions)

1. B,C,D 2. B,D 3. A,B 4. A,B 5. C,D 6. C,D 7. A,B8. A,C 9. B,D 10. B,D 11. B,D 12. B,C

EXERCISE - IIISubjective - type Questions

1. 68°43’ 37.8” 2. 2 10

11 4. a2b2 + 4a2 = 9b2 14. sin

103

2x and cos

101

2x

22. – 41

, 41

23. (i) (i) 2, –1 (ii) 2, 0 27. r1 : r2 = 8 : 5 28. 25

cm

31. (i) 4 (ii) 4 (iii) 4 (iv) 3 32. 1 – 2a2 – 2b2

Comprehension - based Questions

34. C 35. A 36. B 37. C 38. B 39. D

Matrix Match - type Questions

40. (A)–(Q, S) ; (B)–(P) ; (C)–(Q) ; (D)–(P)

EXERCISE - IVPrevious Year’s Question

Level - I(JEE Main)1. B 2. B 3. A 4. D 5. D 6. D7 D

Level - II(JEE Advanced)1 (a) A,B (b) C,D 2. 0002 3. 0007 4. A,C,D 5. 86. C 7. A,B 8. 0.5

7. Theory_Trigo _ I8. Exercise_Trigo_I

chapter · 2020. 9. 26. · chapter 4 section - a angle and its units (1) angle consider a ray oa....

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