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132 INTRODUCTION TO TRIGONOMETRY MATHEMATICS–X

CHAPTER 8INTRODUCTION TO TRIGONOMETRY

Points to Remember :1. If ABC is a right triangle right angled at B and BAC = , 0° 90°, we have :

Base = AB, Perpendicular = BC and, Hypotenuse = AChere,

BC Perpendicularsin =AC Hypotenuse

;AB Basecos =AC Hypotenuse

BC Perpendiculartan =AB Base

;AC Hypotenusecosec =BC Perpendicular

AC HypotenusesecAB Base

;AB BasecotBC Perpendicular

2. We have, 1 1 1cosec , sec , cotsin cos tan

Also, sin costan , cotcos sin

3. Values of various Trigonometric ratios :

T-ratio 0° 30° 45° 60° 90°

sin 012

12

32

1

cos 13

212

12

0

tan 013

1 3 not defined

cosec not defined 2 223

1

sec 123 2 2 not defined

cot not defined 3 113

0

C

B A90°

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MATHEMATICS–X INTRODUCTION TO TRIGONOMETRY 133

4. The value of sin increases from 0 to 1 as increases from 0 ° to 90°. Also, the value of cos decreasesfrom 1 to 0 as increases from 0° to 90°.

5. If is an acute angle, thensin (90° – ) = cos , cos (90° – ) = sin tan (90° – ) = cot , cot (90° – ) = tan sec (90° – ) = cosec , cosec (90° – ) = sec

6. Basic trigonometric identities :

(i) 2 2sin cos 1 or 2 21 cos sin or 2 21 sin cos

(ii) 2 21 tan sec or 2 2sec tan 1

(iii) 2 21 cot cosec or cosec2 – cot2 = 1

ILLUSTRATIVE EXAMPLES

Example 1. In ABC, right angled at B, AB = 24 cm, BC = 7 cm. Determine : (i) sin A, cos A (ii) sin C, cos C [NCERT]

Solution. In right angled ABC, we haveAC2 = AB2 + BC2

AC2 = (24)2 + (7)2

AC2 = 576 + 49 AC2 = 625 AC = 25

(i)opposite side BCsin Ahypotenuse AC

725

adjacent side ABcos Ahypotenuse AC

2425

C

A

B

24 c

m

7cm(ii)

opposite side ABsin Chypotenuse AC

2425

adjacent side BCcosChypotenuse AC

725

Example 2. Given : ,20cot21

find all other trigonometric ratios.

Solution.20 Base 20cot21 Perpendicular 21

Let Base (B) = 20 k, then perpendicular (P) = 21 k

Since P2 + B2 = H2 where H = hypotenuse

P = 21k

B = 20k

H = 29k

H2 = (21 k)2 + (20 k)2

= 441 k2 + 400 k2 = 841 k2

2H 841 29k k

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P 21sinH 29

kk

2129 ;

B 20cosH 29

kk

2029

21tan20

P kB k

2120 ;

29cosec21

H kP k

2921

and 29sec20

H kB k

2920

Example 3. If 3 cot A = 4, check whether 2

2 22

1 – tan A = cos A – sin A1+tan A

or not. [NCERT]

Solution.4 1 3cot A tan A3 cot A 4

consider,

2

2

22

3 91 11 tan A 16 9 74 1691 tan A 16 9 253 11 164

In right ABC, AC2 = AB2 + BC2

= (4k)2 + (3k)2 = 16k2 + 9k2 = 25k2

2AC 25 5k k

Now, AB 4cosAAC 5

kk

and BC 3sin AAC 5

kk

C

3k

BA

5k

4k

2 22 2 4 3cos A sin A

5 5k kk k

2 2 2

2 2

16 9 7 72525 25

k k kk k

from above, we conclude that 2

2 22

1 tan cos sin1 tan

A A AA

.

Example 4. Evaluate : 2 2 2

2 25cos 60° +4sec 30° – tan 45°

sin 30° +cos 30° [NCERT]

Solution.2 2 2

2 2

5cos 60 4sec 30 tan 45sin 30 cos 30

22

2

22

1 25 4 (1)2 3

1 32 2

5 16 15 64 1214 3 12

1 3 44 4 4

6712

Ans.AMIT BAJA

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Example 5. If tan (A + B) = 3 and tan (A – B) 13

; 0° < A + B 90°; A > B, find A and B. [NCERT]

Solution. tan (A + B) = 3 = tan 60° A + B = 60° ...(1)

and, 1tan (A B) tan 30 A B 303

...(2)

On adding (1) and (2), we get, 2A = 90° A = 45°subtracting (2) from (1), we get, 2B = 30° B = 15°So, A = 45° and B = 15° Ans.

Example 6. Evaluate :

(i) cos (40° – ) – sin (50° + ) + 2 2

2 2

cos 40 cos 50sin 40 sin 50

[CBSE 2002]

(ii)2 sin 68 2 cot 15 3 tan 45 tan 20 tan 40 tan 50 tan 70cos 22 5 tan 75 5

[CBSE 2004]

Solution. (i) We have, cos (40° – ) – sin (50° + ) + 2 2

2 2


2 2

2 2

cos 40 cos (90 40 )sin[90 (40 )] sin(50 )sin 40 sin (90 40 )

[ cos sin (90 )]

2 2

2 2

cos 40 sin 40sin (50 ) sin(50 )sin 40 cos 40

101

1 Ans. [ sin2 + cos2 = 1]

(ii) We have, 2sin 68 2cot15 3 tan 45 tan 20 tan 40 tan 50 tan 70cos 22 5 tan 75 5

2sin(90 22 ) 2cot15 3tan 45 tan 20 tan 40 tan (90 40 ) tan (90 20 )

cos 22 5 tan(90 15 ) 5

2 cos 22 2 cot15 3.1.tan 20 tan 40 cot 40 cot 20cos 22 5cot15 5

[ sin (90° – ) = cos , tan (90° – ) = cot ]

2 3 tan 20 tan 402 1 15 5 tan 40 tan 20

1cottan

2 32 1 2 15 5

1 Ans.

Example 7. If tan 2 A = cot (A – 18°), where 2A is an acute angle, find the value of A. [NCERT]

Solution. We have, tan 2A = cot (A – 18°) = tan [90° – (A – 18°)] [ cot tan (90 )] = tan (108° – A)

2A = 108° – A

3A = 108° 108A3

A = 36° Ans.

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Example 8. If A, B, C are the interior angles of a triangle ABC, show that B+C Asin = cos2 2

. [NCERT]

Solution. We know, in any ABC, A + B + C = 180° B + C = 180° – A

B C 180 A A90

2 2 2

taking sine to both sides, we get

B C Asin sin 902 2

B C Asin cos

2 2

[ sin (90 ) cos ]

Hence shown.Example 9. Prove the following Trigonometric identities :

(i)11

cos cosec cotcos

(ii) (sin + cosec )2 + (cos + sec )2 = 7 + tan2 + cot2 [CBSE 2000]

(iii)tan sec 1 1 sintan sec 1 cos

[CBSE 2002]

(iv)1 1 1 1– –

cosec A – cot A sin A sin A co sec A cot A

[CBSE 2002(C)]

(v)tan A cot A 1 tan A cot A 1 sec A cosec A

1 cot A 1 tan A

[CBSE 2002(C)]

Solution. (i) We have, LHS = 1 cos 1 cos 1 cos1 cos 1 cos 1 cos

= 22

2 2

(1 cos )(1 cos )1 cos sin

1 cos 1 cos

sin sin sin

= cosec + cot = RHS(ii) We have, LHS = (sin + cosec )2 + (cos + sec )2

= sin2 + cosec2 + 2 sin cosec + cos2 + sec2 + 2 cos sec

= 2 2 2 2 1 1(sin cos ) cosec sec 2sin 2cossin cos

2 21 cosec sec 2 2

2 2 1 1sin cos 1, cosec , secsin cos

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= 5 + (1 + cot2) + (1 + tan2) [ cosec2 = 1 + cot2, sec2 = 1 + tan2] = 7 + tan2 + cot2 = RHS

(iii) We have, tan sec 1LHStan sec 1

(tan sec ) 1(tan sec ) 1

2 2(sec tan ) (sec tan )

tan sec 1

[ sec2 – tan2 = 1]

(sec tan ) (sec tan )(sec tan )

tan sec 1

(sec tan )[1 (sec tan )]

tan sec 1

(sec tan )(1 sec tan )

tan sec 1

1 sinsec tan

cos cos

1 sin RHS

cos

(iv) We have, 1 1LHScos ec A cot A sin A

1 cosec A cot A 1.

cosecA cot A cosec A cot A sin A

2 2

cosec A cot A cosec Acosec A cot A

= cosec A + cot A – cosec A [ cosec2 – cot2 = 1] = cot A ...(1)

1 1RHS

sin A cosec A cot A

1 1 cos ec A cot A.

sin A cosec A cot A cosec A cot A

2 2

1 cosec A cot Asin A cosec cot A

cosec A (cosec A cot A) 2 2[ cosec cot 1] = cot A ...(2)

from (1) and (2), it follows that LHS = RHS.AMIT B

AJAJ


(v) We have, tan A cot ALHS1 cot A 1 tan A

1tan A tan A

1 1 tan A1tan A

1cot

tan

tan A 1tan A 1 tan A(1 tan A)

tan A

2tan A 1tan A 1 tan A(1 tan A)

2tan A 1tan A 1 tan A (tan A 1)

3tan A 1tan A(tan A 1)

2(tan A 1)(tan A tan A 1)tan A (tan A 1)

[ a3 – b3 = (a – b) (a2 + ab + b2)]

2tan A tan A 1tan A

2tan A tan A 1tan A tan A tan A

= tan A + 1 + cot A = 1 + tan A + cot A

sin A cos A1cos A sin A

2 2sin A cos A1cos Asin A

11 1 cosec A sec A = RHSsin A cos A

Example 10. (i) If tan + sin = m and tan – sin = n, show that 2 2 4m n mn . [CBSE 2000]

(ii) If sec + tan = p, show that 2

2

p 1 sinp 1

Solution. (i) We have, LHS = m2 – n2

= (tan + sin )2 – (tan – sin )2

= (tan2 + sin2 + 2tan . sin ) – (tan2 + sin2 – 2 tan sin )= 4 tan sin ...(1)

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RHS 4 mn

4 (tan sin )(tan sin )

2 24 tan sin

2 2 2 22

2 2

sin sin sin cos4 sin 4cos cos

2 2 2 2

2 2

sin (1 cos ) sin .sin4 4cos cos

4 2

2

sin sin sin4. 4. 4.sin .cos coscos

= 4 sin tan = 4 tan . sin ...(2)

from (1) and (2), it follows LHS = RHS.

(ii) We have, 2 2

2 2

1 (sec tan ) 1LHS1 (sec tan ) 1

pp

2 2

2 2

sec tan 2sec tan 1sec tan 2sec tan 1

2 2

2 2

(sec 1) tan 2sec tansec 2sec tan (1 tan )

2 2

2 2

tan tan 2sec tansec 2sec tan sec

2 2[ 1 tan sec ]

2

2

2 tan 2sec tan2sec 2sec tan

2 tan (tan sec )2sec (sec tan )

tan sin sinsec cos .sec 1

= sin = RHS

PRACTICE EXERCISE

Question based on Trigonometric Ratios :

1. In ABC, right angled at B, if AB = 12 cm and BC = 5 cm, find (i) sin A and tan A (ii) sin C and cot C.2. In each of the following, one of the six trigonometric ratios is given. Find the values of the other

trigonometric ratios.

(i) 15sin17

(ii) 3cot4

(iii) 7cos25

(iv) 13sec5

(v) 41cosec9

(vi) 5tan12

AMIT B

AJAJ


3. If tan A 2 1, show that 2sin A.cos A

4

4. If 12cos A ,13

verify that 35sin A(1 tan A)156

.

5. If 1sin A2

, verify that 3 cos A – 4 cos3 A = 0.

6. If 5sin ,

13 verify that tan2 – sin2 = tan2 . sin2.

7. If 7 cot 24, prove that 1 cos 11 cos 7

.

8. If 5sec ,4

verify that 2

tan sinsec1 tan

9. If 4cot 5, show that 5sin 3cos 75sin 2cos 2

10. If 2 tan = 1, find the value of 3cos 2sin2cos sin

.

11. If 13cot12

, find the value of 2 2

2sin coscos sin

12. If tan ,ab

find the value of cos sincos sin

13. If 13sec5

, show that 3cos 2sin 39cos 4sin

14. In ABC, right angled at B, AC + BC = 25 cm and AB = 5 cm, find the value of sin2A + cos2A.15. If 21 cosec = 29, find the value of :

(i) 2 2

2

cos sin1 2sin

(ii) 2

2 2

2cos 1cos sin

16. If 1tan 2;tan

show that 22

1tan 2tan

17. If 1sin 3,sin

AA

find the value of 22

1sinsin

AA

18. If cot 3, show that :

(i) 2

2

2 cos 1172 sin

(ii) cosec2– cot2 = 1

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19. If 5sec4

verify that 3 3

3 2

3sin 4sin 3tan tan4cos 3cos 1 3tan

20. If B and Q are acute angles such that cos B = cos Q, show that B = Q.

Questions based on Trigonometric ratios of some specific angles

21. Evaluate each of the following : (i) sin 60° cos 30° + cos 60° sin 30° (ii) cos 60° cos 30° – sin 60° sin 30°(iii) 2 cos2 60 cot 30° + 6 sin2 30° cosec2 60° (iv) 4 cot2 45° – sec2 60° + sin2 60° + cos2 90°

(v) 2

2 2

cot 30cos 30 sin 30

(vi) 2 2 25sin 30 cos 45 4 tan 30

2sin30 cos30 tan 45

(vii) 2 (cos2 45° + tan2 60°) – 6 (sin245° – tan230°)

(viii) 2 2 2 2

2

tan 60 3sec 30 4 cos 45 5cos 90cosec 30 sec60 cot 30

22. Verify each of the following :

(i) sin 60° = 2 sin 30° cos 30° 2

2 tan 301 tan 30

(ii)

22 2

2

1 tan 30cos 60 cos 30 sin 301 tan 30

(iii) 2

2 tan 30tan 601 tan 30

23. Show that :

(i) cosec2 30° sin2 45° – sec2 60° + 2 = 0 (ii) 2 2 2 134 tan 30 sec 30 sin 456

(iii) 2

2

cos 60 3cos 60 2 1sin 60

(iv) 4 (sin2 30° + cos2 60°) – 3 (cos2 45° – tan245°) = 72

24. If = 30°, verify that :

(i) 2

2 tansin 21 tan

(ii)

2

2

1 tancos21 tan

(iii) 2

2 tantan 21 tan

(iv) sin 3 = 3 sin – 4 sin3

25. If A = 45°, verify that :(i) sin 2 A = 2 sin A . cos A (ii) cos 2 A = 2 cos2 A – 1

(iii) 1 cos 2Asin A

2

(iv) 1 cos 2Acos A2

26. If A= 60° and B = 30°, verify that :(i) sin (A + B) = sin A cos B + cos A sin B (ii) sin (A – B) = sin A cos B – cos A sin B(iii) cos (A + B) = cos A cos B – sin A sin B (iv) cos (A – B) = cos A cos B + sin A sin B

(v) tan A tan Btan(A B)1 tan A.tan B

(vi) tan A tan Btan(A B)

1 tan A.tan B

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27. Given that sin (A + B) = sin A cos B + cos A sin B, find the value of sin 75°.28. Given that cos (A – B) = cos A cos B + sin A sin B, find the value of cos 15°.29. Find x in each of the following, if :

(i) 2 cos x = 1 (ii) 2 sin 2x = 1

(iii) tan 3 3x (iv) 2sin 3 3x

(v) 3 cot 3 1x (vi) 3.sec 22x

30. Find acute angle in each of the following cases; if,

(i) sin (3 – 15°) = 1 (ii) 2 sin (3 – 15°) 3

(iii) 2 1sin4

(iv) 23tan 1 0

(v) cos (40° + ) = sin 30° (vi) 2cot (2 5 ) 3

31. If 3sin (A 2B)

2 and cos (A + 4B) = 0, find the values of angles A and B.

32. If sin (A + B) = 1 and 3cos (A B) ; 0 A B 90 ,

2 and A > B; find A and B.

33. If sin (A – B) 12

and cos (A + B) 12

; 0 < A + B 90°, and A > B, find A and B.

34. ABC is a right triangle, right angled at C. If A = 30°, and AB = 40 units, find the remaining two sides andB of ABC.

35. ABCD is a rectangle with AD =12 cm and DC = 20 cmas shown. The line segment DE is drawn making anangle of 30° with AD, intersecting AB in E. Find thelengths of DE and AE.

Questions based on Trigonometric Identitities of Complementary Angles

36. Evaluate each of the following :

(i) cos 49sin 41

(ii)

sec32cosec 58

(iii)

tan 21cot 69

37. Evaluate each of the following :(i) sin 54° – cos 36° (ii) tan 62° – cot 28°(iii) cosec 47° – sec 43° (iv) sec2 31° – cot2 59°(v) sin2 29° + sin2 61 (vi) tan2 48° – cosec242°


(i) cos10 cos 59 cosec 31sin 80

(ii)

tan 53 cot 79cot 37 tan11

E B

C

A

D 20 cm

12 c

m

30°

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(iii) 3sin 62 sec 42cos 28 cosec 48

(iv) 2cos 67 tan 40 sin 90sin 23 cot 50


(i) 2 2

2 2


(ii) 2 2sin 27 cos63

cos63 sin 27

(iii) 2 2tan 36 cot 54

cot 54 tan 36

(iv) 22cosec 27 sec 63

sec 63 c osec 27

40. Evaluate each of the following :(i) tan 5° tan 25° tan 30° tan 45° tan 65° tan 85° (ii) cot 12° cot 38° cot 52° cot 60° cot 78°(iii) sec (35° + A) – cosec (55° – A) (iv) cos (36° + A) – sin (54° – A)

41. Prove that :

(i) cos (90 ) sin 2

sin cos (90 )

(ii) cos . cos (90° – ) + cos . sin (90° – ) = 1

(iii) sin .cos (90 cos cos sin (90 ).sin 1sin (90 ) cos (90 )

(iv) cos (90 ) . sec (90 ) . tan tan (90 ) 2cos ec (90 ) sin (90 ) .cot (90 ) cot

42. Without using trigonometric tables, find the value of each of the following :(i) sin (50° + ) – cos (40° – ) + tan 1° tan 10° tan 20° tan 70° tan 80° tan 89° [CBSE 2002]

(ii) 2 2

2 2

cos 40 cos 50cos (40 ) sin (50 )sin 40 sin 50

[CBSE 2002]

(iii) 2 2

2 2 2sin 47 cos 43 4cos 45 cosec 67 tan 23cos 43 sin 47

(iv) 2 2 sin15 cos 75 cos15 sin 75sec 10 cot 80cos sin(90 ) sin . cos (90 )

[CBSE 2002]

(v) cot tan (90° – ) – sec (90° – ) cosec + sin2 25° + sin2 65° + 3 (tan 5° tan 45° tan 85°)

(vi) 22 2

3tan 25 tan 40 tan 50 tan 65 1 tan 6024 (cos 29 cos 61 )

[CBSE 2004]

(vii) 2 2tan cot (90 ) sec cosec (90 ) sin 35 sin 55

tan10 tan 20 tan 45 tan 70 tan 80

[CBSE 2002 (C), 2005]

(viii) 2 2 2 2 2

2 2 2 2

cosec (90 ) tan 2 tan 30 sec 52 sin 324 (cos 48 cos 42 ) cosec 70 tan 20

[CBSE 2006]AMIT B

AJAJ


(ix) 2 2tan 20 cot 20 2 tan15 tan 37 tan 53 tan 60 tan 75

cosec70 sec70

[CBSE 2003]

(x) 2 2sec39 2 .tan17 tan 38 tan 60 tan52 tan 73 3(sin 31 sin 59 )cosec 51 3

[CBSE 2006 (C)]

43. Express each of the following in terms of trigonometric ratios of angles between 0° and 45° :(i) sin 79° + cosec 83° (ii) tan 58° + sec 46°(iii) cosec 57° + cot 57° (iv) sin 82° + tan 82°

44. In any ABC, prove that :

(i) A B Csin cos

2 2

(ii)

B C Atan cot2 2

45. If sin 3 = cos ( – 6°), where 3 and – 6°) are acute angles, find the value of .46. If tan 2 = cot ( + 6°), where 2 and + 6° are acute angles, find the value of .47. If sec 5 = cosec ( – 36°), where 5 is an acute angle, find the value of .48. If sin A = cos B, prove that A + B = 90°.49. Prove that :

(i) tan 1° tan 2° tan 3° ..... tan 89° = 1

(ii) sin2 5 + sin2 10° + sin2 15° + ..... + sin2 85° + sin2 90° 19 .2

50. If A and B are acute angles and sin (A + B) = cos (A – B), show that A = 45°.

Questions based on Trigonometric Identities :

51. Using trigonometric identities, write the following expressions as an integer :(i) 5 cot2 A – 5 cosec2 A (ii) 4 tan2 – 4 sec2 (iii) 7 cos2 + 7 sin2 (iv) 3 sec2 – 3 tan2

52. Simplify the following expressions :

(i) (1 + cos ) (cosec – cot ) (ii) 3 3sin cos

sin cos

(iii) 1 1

1 sin 1 sinA A

(iv)

4 4

2 2

sin cossin cos

A AA A

(v) 2

2

1 tancot 1

(vi) cosec (1+ cos ) (cosec – cot )

Prove that following identities (53–100) :

53. cos2 (1 + tan2) = 1 54. cosec2 + sec2 = cosec2 . sec2 [CBSE 2001]

55.2 2sin cos tan cot

sin cos

56. 2 1 cos(cot cosec )1 cos

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57. (1 + tan2 ) (1 + sin ) (1 – sin ) = 1 58. sin A cosec A cot A

1 cos A

59.22sin 1tan cot

sin cos

60. 21 1 2cosec1 cos 1 cos

61.tan sin sec 1tan sin sec 1

62. sec A (1 – sin A) (sec A + tan A) = 1

63.1 sin cos 2sec

cos 1 sin

64. 22

2

1 tan 1 tan1 cot1 cot

65.tan cot 1 tan cot 1 sec cosec

1 cot 1 tan

[CBSE 2002 (C)]

66.21 cos 1 cos

sin 1 cos

67. 1 cot A sin A cos A1 cot A sin A cosA

68.2

24

1 cot sin1 cot

69. 1 sin sec tan1 sin

A A AA

70.1 sin cos1 sin 1 sin

A AA A

71.

1 cos cosec cot1 cos

72.1 cos cosec cot1 cos

73. 1 cosec cot

cosec cot

74.1 cos 1 cos 2 cosec1 cos 1 cos

75. (sin + sec )2 + (cos + cosec )2 = (1 + sec cosec )2

76. 2 2 2

sin A cos A sin A cos A 2 2sin A cos A sin A cos A sin A cos A 2sin A 1

[CBSE 2000 (C)]

77.3

3

sin 2sin tan2cos cos

[CBSE 2000] 78.

sin sin2cot cosec cot cosec

[CBSE 2000]

79. (1 + cot – cosec ) (1 + tan + sec ) = 2 [CBSE 2000]

80.2cosec cosec 2sec

cosec 1 cosec 1

81. 2 2 2 4

1 1 11 1tan A cot A sin A sin A

[CBSE 2006(C), 2007]

82.1 1 1 1

sec A tan A cos A cos A sec A tan A

[CBSE 2005]

83.1 1 1 1

cosec A cot A sin A sin A cosec A cot A

[CBSE 2002C, 2006]AMIT B

AJAJ


84.sec A tan A 1 1 sin Atan A sec A 1 cos A

85. cot A cosec A 1 1 cos Acot A – cosec A + 1 sin A

86.sin cos 1 1 sinsin cos 1 cos

[CBSE 2001(C)]

87.sec tan 1 sec tansec tan sec tan

88.sec tan 1sec tansec tan sec tan

89. sec4 – sec2 = tan4 + tan2

90.2 3cos sin 1 sin cos

1 tan sin cos

[CBSE 2000(C)]

91.21 cos sin cot

sin (1 cos )

[CBSE 2003]

92. 2

tan tan 11 cot 2 cosec

93. 2 2sec A cosec A tan A cot A

94. 2 sec2 – sec4 – 2 cosec2 + cosec4 = cot4 – tan4 [CBSE 2000]

95.2 2 2 2

2 22 2 2 2

cos B cos A sin A sin Btan A tan Bcos B.cos A cos A cos B

[CBSE 2005]

96. 2 (sin6 + cos6 ) – 3 (sin4 + cos4) + 1 = 0

97.sec 1 sec 1 2 cosecsec 1 sec 1

[CBSE 2001, 2006(C)]

98.sin sin cos cos 0cos cos sin sin

99.

2 3sin cos 1 sin cos1 cot cos sin

100.sin A cos A 1

sec A tan A 1 cosec A cot A 1

101. If cos + sin = 2 cos , show that cos – sin = 2 sin [CBSE 2002(C)]

102. If sin + cos = p and sec + cosec = q, show that 2( 1) 2q p p

103. If x = a sec + b tan and y = a tan + b sec , prove that x2 – y2 = a2 – b2. [CBSE 2001,2000 (C)]

104. If cos cosand ,cos sin

m n

show that (m2 + n2) cos2 = n2

105. If a cos + b sin = m and a sin – b cos = n prove that a2 + b2 = m2 + n2.AMIT BAJA

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106. If a cos = x and b cot = y, show that 2 2

2 2 1a bx y

107. If cos sinx y ma b

and sin cos ,x y na b

prove that 2 2

2 22 2 .x y m n

a b

108. If 1sec ,4

xx

prove that sec + tan = 2x or 12x [CBSE 2001]

109. If sec tan ,p prove that 2

2

1 sin1

pp

[CBSE 2004]

110. If tan + sin = m and tan – sin = n, prove that 2 2 4m n mn . [CBSE 2000, 2002(C), 2004(C)]

111. If a cos – b sin = c, prove that 2 2 2sin cosa b a b c [CBSE 2001(C)]

112. If 3sin 5cos 5, prove that 5sin 3cos 3

113. If sin + sin2 = 1, prove that cos2+ cos4 = 1 [CBSE 2002(C)]114. 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), prove

that each side is equal to ± 1.115. If x = r sin cos , y = r sin sin and z = r cos , then show that x2 + y2 + z2 = r2.

HINTS TO SELECTED QUESTIONS

9. here, 5cot .4

Consider LHS = 5sin 3cos5sin 2 cos

dividing numerator and denominator by sin , we get

LHS

55 35 3cot 74

55 2cot 25 24

16.1tan 2.

tan

Squaring both sides,

22

1 1tan 2.tan . 4tantan

22

1tan 4 2 2.tan

20. Consider two right triangles ACB and PRQ.

We have, BC QRcos B ,cos QAB PQ

B

A

C Q

P

RSince, BC QRcos B cosQ ,AB PQ

k say ...(1)AMIT BAJA

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Now, 2 2 2 2 2

2 2 2 2 2

AC AB BC AB AB ABPR PQPQ QR PQ PQ

kk

Thus, In ACB and PRQ, we have

AC AB BC ACB PRQ B Q.PR PQ QR

27. sin 75° = sin (45° + 30°) = sin 45° cos 30° + cos 45° sin 30°

31.3sin (A 2B) sin 60 A 2B 60

2 ...(1)

and, cos (A + 4B) = 0 = cos 90° A + 4B = 90° ...(2)solve (1) and (2) now.

49. (ii) sin2 5° + sin2 10° + sin2 15° + .... + sin2 85° + sin2 90°2 2 2 2 2 2(sin 5 sin 85 ) (sin 10 sin 80 ) ... (sin 40 sin 50 ) 2 2sin 45 sin 90

2 2 2 2(sin 5 cos 5 ) (sin 10 cos 10 ) 2 2 2 2... (sin 40 cos 40 ) sin 45 sin 90

22

8 times

11 1 ..... 1 (1)2

1 198 12 2

74. LHS 1 cos 1 cos 1 cos 1 cos1 cos 1 cos 1 cos 1 cos

2 2

2 2

(1 cos ) (1 cos ) 1 cos 1 cossin sin1 cos 1 cos

2 2cosec RHSsin

86. LHS sin cos 1sin cos 1

dividing numerator and denominator by cos ,

LHS tan 1 sec tan sec 1tan 1 sec tan sec 1

2 2(tan sec ) (sec tan )tan sec 1

(tan sec ) (sec tan )(sec tan )tan sec 1

(tan sec )(1 sec tan )tan sec 1

= tan + sec sin 1

cos cos

1 sin RHScos

90. LHS = 2 3cos sin

1 tan sin cos

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2 3 2 3cos sin cos sinsin sin cos cos sin cos sin1cos

3 3 2 2cos sin (cos sin )(cos sin cos sin )cos sin cos sin

= 1 + sin cos = RHS

96. 6 6 2 3 2 3sin cos (sin ) (cos )

2 2 2 2 2 2 2 2(sin cos ) [(sin ) (cos ) (sin )(cos )]

3 3 2 2[ ( )( )]a b a b a b ab

2 2 2 2 2 2 2 21.[(sin ) (cos ) 2sin cos 3sin cos ]

2 2 2 2 2(sin cos ) 3sin cos 2 21 3sin cos

also, sin4 + cos4 = (sin2 )2 + (cos2 )2 + 2 sin2 cos2 – 2 sin2 cos2

= (sin2 + cos2 )2 – 2 sin2 cos2 = 1 – 2 sin2 cos2

Now, put these values in LHS and simplify.

101. cos + sin = 2 cos sin = ( 2 1) cos 1cos .sin

2 1

1 2 1cos sin . cos ( 2 1).sin

2 1 2 1

cos 2 sin sin cos sin 2 sin

108.2 21 1 1sec tan 1

4 4 4x x x

x x x

1tan

4x

x

Now, when 1tan4

xx

, then sec + tan 1 1 24 4

x x xx x

and, when 1 1 1 1tan , then sec tan4 4 4 2

x x xx x x x

111. Consider, (a cos – b sin )2 + (a sin + b cos )2

2 2 2 2 2 2 2 2cos sin 2 sin cos sin cos 2 sin cosa b ab a b ab

2 2 2 2 2 2 2 2(cos sin ) (cos sin )a b a b

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

2 2 2sin cosa b a b c

114. We have, (sec A + tan A) (sec B + tan B) (sec C + tan C) = (sec A – tan A) (sec B – tan B) .(sec C – tan C). Multiplying both sides by (sec A – tan A) (sec B – tan B) (sec C – tan C), we get

(sec2A – tan2A) (sec2B – tan2B) (sec2C – tan2C) = [(sec A – tan A) (sec B – tan B) (sec C – tan C)]2

(sec A – tan A) (sec B – tan B) (sec C – tan C) = ± 1

Similarly, multiplying by (sec A + tan A) (sec B + tan B) (sec C + tan C), we get

(sec A + tan A) (sec B + tan B) (sec C + tan C) = ± 1.

115. LHS = x2 + y2 + z2

= r2 sin2 cos2 + r2 sin2 sin2 + r2cos2

= 2 2 2 2 2 2sin (cos sin ) cosr r

2 2 2 2 2 2 2 2sin cos (sin cos ) RHS.r r r r

MULTIPLE CHOICE QUESTIONS

Mark the correct alternative in each of the following :

1. If 3 tan = 4, then the value of 1 sin1 sin

is :

(a) 1 (b) 12

(c) 13 (d)

23

2. If 5 sin – 3 = 0, then the value of cosec cot2cot

is :

(a) 14

(b) 18 (c)

34

(d) 78

3. If tan ,ab

then the value of sin cossin cos

a ba b

is

(a) 2 2

2 2

a ba b

(b) 2 2

2 2

a ba b

(c) a ba b

(d) –a b

a b

4. If 5 tan 4 0 , then the value of 5sin 3cos5sin 2cos

is

(a) 23 (b)

13 (c)

16 (d)

52

5. If 4sin ,5

then the value of 4 tan 5cossec cot

is

(a) 1 (b) 58 (c)

74

(d) 2829

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6. The value of 4 cot245° – sec260° + sin260° + cos290° is

(a) 14

(b) 24

(c) 34

(d) 1

7. If tan 3x = sin 45° cos 45° + sin 30°, then the value of x is :(a) 15° (b) 30° (c) 45° (d) 60°

8. The value of 4 (sin4 30° + cos4 30°) – 3 (sin2 45° – 2 cos2 45°) is(a) 0 (b) 1 (c) 4 (d) none of these

9. The value of 2 2

2 2

sin 63 sin 27cos 17 cos 73

is

(a) 0 (b) 1 (c) 2 (d) none of these10. The value of tan 5° tan 10° tan 15° tan 75° tan 80° tan 85° is :

(a) –1 (b) 0 (c) 1 (d) none of these

11. The value of sin 75 sin12 cos18 cosec 72cos15 cos78

is :

(a) –1 (b) 0 (c) 1 (d) none of these

12. The value of 2 2sin 35 cos55 2 cos60

cos55 sin 35

is :

(a) 0 (b) 1 (c) –1 (d) 2

13. The value of sin 75 sin15cos15 cos75

is :

(a) 0 (b) –1 (c) 1 (d) none of these

14. If sin cos4 , then the value of is :

(a) 18° (b) 36° (c) 45° (d) none of these

15. The value of 2 2sec cosec (90 ) tan cot (90 ) (sin 35 sin 55 )

tan10 tan 20 tan 45 tan 70 tan 80

is

(a) 1 (b) –1 (c) 2 (d) 316. (cosec A – sin A) (sec A – cos A) (tan A + cot A) is equal to :

(a) 1 (b) –1 (c) 2 (d) –2

17.cosec 1 cosec 1cosec 1 cosec 1

is equal to :

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

cos (d) 2

cos18. If x = r sin cos , y = r sin sin and z = r cos , then

(a) 2 2 2 2x y z r (b) 2 2 2 2x y z r

(c) 2 2 2 2x y z r (d) 2 2 2 2z y x r

19. If a cot + b cosec = p and b cot + a cosec = q, then p2 – q2 is equal to :(a) a2 – b2 (b) b2 – a2 (c) a2 + b2 (d) a + bAMIT B

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20. If (1 + cos ) (1 + cos ) (1 + cos ) = (1 – cos ) (1 – cos ) (1 – cos ), then each of the expression is equalto :(a) sin sin sin (b) cos cos cos (c) ± sin sin sin (d) ± cos cos cos

VERY SHORT ANSWER TYPE QUESTIONS (1 MARK QUESTIONS)

1. What is the maximum value of 1

cosec ?

2. If 3sin 2 ,

2 find the value of cos .

3. If sin cos 1, the find the value of sin cos .

4. If cosec = 157 and 90 , find the value of sec .

5. What is the value cos 1° cos 2° cos 3° .... cos 89° cos 90° ...... cos 180°?6. If 7 sin2 + 3 cos2 = 4, find the value of tan if 0° 90°.7. Write the value of sin2 68° + sin2 22° – 1.8. If A and B are acute angles and sin B = cos A, then write the value of A + B.9. Write the value of sin (55° + ) – cos (35° – ).

10. Express tan 87° + sin 63° in terms of trigonometric ratios of angles between 0° and 45°.11. Write the value of tan 5° tan 35° tan 45° tan 55° tan 85°.

12. Write the simplest form of sin cos

sec (90 ) cosec (90 )

.

13. If tan ( – 36°) = cot 2; 2 and ( – 36°) are acute angles, then find .14. If 2cosec (1 + cos ) (1 – cos ) = k, then find the value of k.15. If sin 2 = cos 3, then find the value of .

16. Find the value of 3

3

cos cos .sin sin

17. If 1tan A3

and 1sin B ,2

find the value of A + B.

18. If cos 3 = 1, then find the value of tan .

19. If A, B and C are the angles of a triangle, then find the value of B Ccot2

in terms of A.

20. If tan2 – 5 tan + 1 = 0, find the value of tan + cot .

PRACTICE TESTM.M : 30 Time : 1 hour

General Instructions :

Q. 1-4 carry 2 marks, Q. 5-8 carry 3 marks and Q. 9-10 carry 5 marks each.AMIT B

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1. If 3 cot = 2, find the value of 4sin 3cos2sin 6cos

.

2. Evaluate : 4 cot2 45° + sin2 60° + cos2 90° – sec2 60°.

3. If sin ( + 36°) = cos , where + 36° and are acute angles, find the value of .4. If cosec + cot = a and cosec – cot = b, prove that ab = 1.5. In OPQ, right angled at P, OP = 7 cm, OQ – PQ = 1 cm. Determine the values of sin Q and cos Q.

6. If = 30°, verify that 2

2

1 tancos 21 tan

.

7. Without using trigonometric tables, evaluate :2 2

2 2

sin 20 sin 70 sin (90 )sin cos (90 ).costan cotcos 20 cos 70

8. Prove that :

tan A cot A 1 sec A . cosec A1 cot A 1 tan A

9. A contractor plans to instal two slides for the children to play in a park. For the children below the ageof 6 years, he prefers to have a slide, which is inclined to an angle of 30° to the ground, whereas for elderchildren, he wants to have a steep slide, inclined at an angle of 60° to the ground. The length of the slidein both cases is 3m. What should be the required length of the ladder, in each case, to reach the top ofthe slide and at what distance should it be placed from the base of the slide?

10. If tan A + sin A = m and tan A – sin A = n, prove that (m2 – n2)2 = 16 mn.

ANSWERS OF PRACTICE EXERCISE

1. (i) 5 5,

13 12 (ii) 12 5,13 12

2. (i) 8 15 17 17 8cos , tan , cosec , sec ,cot

17 8 15 8 15

(ii) 4 3 4 5 5sin , cos , tan ,cosec , sec5 5 3 4 3

(iii) 24 24 25 25 7sin , tan , sec ,cosec ,cot25 7 7 24 24

(iv) 12 5 12 13 5sin , cos , tan ,cosec , cot13 13 5 12 12

(v) 9 40 9 41 40sin ,cos , tan , sec , cot41 41 40 40 9

(vi) 5 12 13 13 12sin , cos , cosec , sec ,cot13 13 5 12 5

AMIT B

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10.83 11.

31225 12.

b ab a 14. 1 15. (i) 1 (ii) 1

17. 11 21. (i) 1 (ii) 0 (iii) 3 42

(iv) 34

(v) 6 (vi) 5 (2 3)6

(vii) 6 (viii) 9

27.3 1

2 2

28. 3 1

2 2

29. (i) 60° (ii) 15° (iii) 20° (iv) 20° (v) 20° (vi) 60°

30. (i) 35° (ii) 25° (iii) 30° (iv) 30° (v) 20° (vi) 17.5° 31. A = 30°, B = 15°

32. A = 60°, B = 30° 33. A = 45°, B = 15° 34. AC = 20 3 units, BC = 20 units and B = 60°

35. DE = 8 3 cm, AE = 4 3 cm 36. (i) 1 (ii) 1 (iii) 1

37. (i) 0 (ii) 0 (iii) 0 (iv) 1 (v) 1 (vi) –1 38. (i) 2 (ii) 0 (iii) 2 (iv) 0

39. (i) 1 (ii) 2 (iii) 2 (iv) 0 40. (i) 13

(ii) 13

(iii) 0 (iv) 0

42. (i) 1 (ii) 1 (iii) 1 (iv) 2 (v) 3 (vi) 3

4

(vii) 2 (viii) 5

12

(ix) 1 2 3 (x) 0

43. (i) cos 11° + sec 9° (ii) cot 32° + cosec 44° (iii) sec 33° + tan 33° (iv) cos 8° + cot 8°

45. 24° 46. 28° 47. 21° 51. (i) –5 (ii) –4 (iii) 7 (iv) 3

52. (i) sin (ii) 1 – sin cos (iii) 2 sec2 A (iv) 1 (v) tan2 (vi) 1

ANSWERS OF MULTIPLE CHOICE QUESTIONS

1. (c) 2. (b) 3. (b) 4. (c) 5. (d)6. (c) 7. (a) 8. (c) 9. (b) 10. (c)

11. (c) 12. (b) 13. (c) 14. (a) 15. (c)16. (a) 17. (d) 18. (a) 19. (b) 20. (c)

ANSWERS OF VERY SHORT ANSWER TYPE QUESTIONS

1. 1 2. 3

23. 0 4. 15

75. 0

6.13

7. 0 8. 90° 9. 0 10. cot 3° + cos 27°

11. 1 12. 1 13. 42° 14. 1 15. 18°

16. tan 17. 75° 18. 0 19. Atan2

20. 5

ANSWERS OF PRACTICE TEST

1.13 2.

34

3. 27° 5. 7 24sin , cos25 25

7. 2 9. 2 3 m, 3 m

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