RD Sharma Solutions Chapter 6 Trigonometric Identities Exercise 6.1 Class 10 Maths

RD Sharma Solutions Chapter 6 Trigonometric Identities Exercise 6.1 Class 10 Maths

Chapter Name

RD Sharma Chapter 6 Trigonometric Identities

Book Name

RD Sharma Mathematics for Class 10

Other Exercises

  • Exercise 6.2

Related Study

NCERT Solutions for Class 10 Maths

Exercise 6.1 Solutions

Prove the following trigonometric identities : 

1. ( 1 – cos2 A) cosec2 A = 1 

Solution


2. (1 + cos2 A)sin2 A = 1 

Solution


3. tan2 θ cos2 θ = 1 - cos2 θ

Solution 


4. cosec θ √(1 - cos2 θ) = 1 

Solution 


5. (sec2 θ - 1)(cosec2 θ - 1) = 1 

Solution

We know that sec2 θ - tan2 θ = 1 
⇒ sec2 θ = 1, tan θ 
cosec2 θ - cos2 θ = 1 
cosec2 θ - cot2 θ 
tan2 θ .cot2 θ = tan2 θ(1/tan2 θ)


6. tanθ(1/tanθ) = secθ cosecθ

Solution 


7. cosθ/(1 - sinθ) = (1 + sinθ)/cos θ 

Solution


8.  cos θ/(1 + sinθ) = (1 - sin θ)/cos θ

Solution


9. cos2 A + 1/(1 + cot2 A) = 1 

Solution

1 + cot2 A = cosec2 A  [∵ cosec2 A - cot2 A = 1, cosec2 A = 1 + cot2 A]
⇒ cot2 A + 1/cosec2 A
⇒ cos2 A + sin2 A = 1 
∴ LHS = RHS


10. sin2 A + 1/(1 + tan2 A) = 1 

Solution 

1 + tan2 A = sec2 A    [∵ sec2 A - tan2 A = 1]
⇒ sin2 A + 1/sec2 A  [1 + tan2 A - sec2 A]
⇒ sin2 A + cos2 A = 1 
∴ LHS = RHS


11. √(1 - cos θ)/(1 + cos θ) = cosec θ - cot θ. 

Solution


12. (1 - cosθ)/sinθ = sinθ/(1 + cosθ)

Solution


13. sin θ/(1 - cos θ) - cosec θ + cot θ

Solution


14. (1 - sin θ)/(1 + sin θ) - (sec θ - tan θ)2 

Solution


15. (cosec θ + sin θ)(cosec θ - sin θ) = cos2 θ

Solution  


16. (1 + cot2 θ)tanθ/sec2 θ = cot θ

Solution


17. (sec θ + cos θ)(sec θ - cos θ) = tan2 θ + sin2 θ

Solution


18. sec A(1 - sin A)(sec A + tan A) = 1 

Solution


19. (cosec A - sin A)(sec A - cos A)(tan A + cot A) = 1 

Solution


20. tan2 θ - sin2 θ = tan2 θ .sin2 θ

Solution


21. (1 + tan2 θ)(1 - sin θ).(1 + sin θ) = 1 

Solution


= 1 
= LHS = RHS  Hence proved 

22. sin2 A cot2 A + cos2 A tan2 A = 1 

Solution


23. (i) cos θ - tan θ = (2cos2 θ - 1)/(sin θ cos θ) 

Solution


(ii) tan θ - cot θ = (2 sin2 θ - 1)/(sin θ  cos θ)

Solution


24. cos2 θ/sin θ - cosec θ + sin θ = θ

Solution


25. 1/(1+sinA) + 1/(1 - sinA) = 2sec2 A

Solution


26. (1 + sin θ)/cos θ + cos A/(1 + sin θ) = 2 sec θ

Solution 


27. ([1 + sin θ)2  + (1 - sin θ)2 ]/2 cos2 θ = (1 + sin2 θ)/(1 - sin2 θ)

Solution


28. (1 + tan2 θ)/(1 + cot2 θ) - [(1 - tan θ)/cotθ]2 - tan2 θ

Solution


29. (1 + sec θ)/sec θ = sin2 θ/(1 - cosθ)

Solution


30. tan θ/(1- cotθ) = cotθ/(1 - tan θ) = 1 + tan θ + cot θ.

Solution

 

31. sec6θ = tan6θ + 3tan2 θ sec2 θ + 1 

Solution

We know that sec2 θ - tan2 θ : 
Cubing on both sides 
(sec2  θ - tan2 θ)3 = 1
sec6 θ - tan6 θ - 3sec2 θ.tan2 θ(sec2 θ - tan2 θ) = 1
[∵(a - b)3  = a3 - b3 - 3ab(a - b)]
⇒ sec6 θ - tan6 θ =  3sec2 θ.tan2 θ = 1
⇒ sec6 θ = tan6 θ+ 3sec2 θ.tan2 θ + 1
Hence, proved.


32. cosec6 θ = cot6 θ+ 3cot2 θ cosec2 θ + 1 

Solution 

We know that cosec2 θ - cot2 θ = 1 
Cubing on both sides 
(cosec2 θ - cot2 θ)3 = (1)3 


33. (1+ tan2 θ)cot θ/ cosec2 θ = tan θ
Solution

34. (1 + cos A)/sin2 A = 1/(1 - cos A)
Solution
We know that sin2 A + cos2 A = 1
Sin2 A = 1 - cos2 A 
⇒ sin2 A = (1 - cos A)(1 + cos A)

35. (sec A - tan A)/(sec A + tan A) = cos2 A/(1 + sin A)2 
Solution
LHS = (sec θ - tan θ)/(sec A + tan A)
Rationalizing the denominator, multiply and diving with sec A + tan A we get 

36. (1 + cos A)/sin A = sin A/(1 - cos A)
Solution

37. √(1 + sin A)/(1 - sin A) = sin A + tan A
Solution

38. √(1 - cos A)/(1 + cos A) = cosec A - cot A
Solution

39. (sec A - tan A)2 = (1 - sin A)/(1 + sin A)
Solution

40. (1 - cos A)/(1 + cos A) = (cot A - cosec A)2 
Solution


41. 1/(sec A - 1) + 1/(sec A + 1) = 2cosec A. cot A
Solution

42. cos A/(1 - tan A) + sin A/(1 - cot A) = sin A + cos A
Solution

43. cosec A/(cosec A - 1) + cosec A/(cosec A + 1) = 2sec2 A.
Solution

44. (1 + tan2 A) + (1 + 1/tan2 A) = 1/(sin2 A - sin4 A)
Solution

45. tan2 A/(1 + tan2 A) + cot2 A/(1 + cot2 A) = 1 
Solution


46. (cot A - cos A)/(cos A + cos A) = (cosec A - 1)/(cosec A + 1)
Solution

47. (i) (1 + cos θ + sin θ)/(1 +  cos θ - sin θ)
(ii) (sin θ - cosθ + 1)/(sin θ + cos θ - 1) 
(iii) (cos θ - sin θ + 1)/(cos θ + sin θ - 1) = cosec θ + cot θ
Solution
(i) (1 + cos θ + sin θ)/(1 +  cos θ - sin θ)

(ii) (sin θ - cosθ + 1)/(sin θ + cos θ - 1)

(iii) (cos θ - sin θ + 1)/(cos θ + sin θ - 1) = cosec θ + cot θ

48. 1/(sec A + tan A) - 1/(cos A) = 1/cos A - 1/(sec A - tan A) 
Solution

49. tan2 A + cot2 A = sec2 A cosec2 A - 2
Solution

50. (1 - tan2 A)/(cot2 A- 1) = tan2 A.
Solution

51. 1 + cot2 θ/(1 + cosec θ) = cosec θ
Solution

52. cos θ/(cosecθ + 1) + cos θ/(cosec θ -1) = 2 tan θ
Solution

53. (1 + cos θ - sin2 θ)/[sin θ(1 + cos θ)] = cot θ
Solution

54. tan3 θ/(1 + tan2 θ) + cot3 θ/(1 + cot2 θ) = sec θ cosec θ - 2sin θ cos θ
Solution

55. If  Tn = sinn θ + cosn θ, prove that (T3 – T5 )/T1 = (T5 – T7)/T3
Solution

56.  [tan θ + 1/cos θ]2 + [tan θ - 1/cos θ]2  = 2[(1 + sin2 θ)/(1 - sin2 θ)
Solution 

57. [1/(sec2 θ – cos2 θ) + 1/(cosec2 θ – sin2 θ)]sin2 θcos2 θ = (1 – sin2 θcos2 θ)/(2 + sin2 θcos2 θ)
Solution

58. [(1 + sin θ - cos θ)/(1 + sin θ + cos θ)]2 = (1 - cos θ)/(1 + cos θ)
Solution

59. (sec A + tan A - 1)(sec A - tan A + 1) = 2tan A 
Solution

60.  (1 + cot A - cosec A)(1 + tan A + sec A) = 2 
Solution

61. (cosec θ - sec θ)(cot θ - tan θ)(cosec θ + sec θ)(sec θ cosec θ - 2)
Solution

62. (sec A - cosec A)(1 + tan A + cot A) = tan A sec A - cot A cosec A
Solution

63. (cos A cosec A - sin A sec A)/(cos A + sin A) = cosec A - sec A 
Solution

64. sin A/(sec A + tan A -1) + cot A/(cosec A + cot A - 1) = 1
Solution

65. tan A/(1 + tan2 A) + cos A/(1 + cot2 A)2 = sin A cos A
Solution 

66. sec4 A(1 - sin4 A) - 2 tan4 A = 1 
Solution

67. cot2 A(sec A - 1)/(1 + sin A) = sec2 [(1 - sin A)/(1 + sec A)]
Solution

68. (1 + cot A + tan A)(sin A - cos A) = sec A/(cosec2 A) - cosec A/sec2 A = sin A tan A - cos A cot A
Solution

69.  sin2 A cos2 B - cos2 A sin2 B = sin2 A - sin2 B
Solution

70. (cot A + tan B)/(cot B + tan A) = cot A tan B
Solution

71. (tan A + tan B)/(cot A + cot B) = tan A  tan B
Solution

72. cot2 A cosec2 B - cot2 B cosec2 A = cot2 A - cot2 B
Solution

73. tan2 A sec2 B - sec2 A tan2 B = tan2 A - tan2 B
Solution

74. If x = a sec θ + b tan θ and y = a tan θ + b sec θ, prove that x2 - y2  = a2 - b2 
Solution

75. If x/a cos θ + y/b sin θ = 1 and x/a sin θ - y/b cos θ = 1, prove that x2/a2 + y2/ b2  = 2 
Solution

76. If cosec θ - sin θ = a2 , sec θ - cos θ = b3 , prove that a2b2(a2 + b2 ) = 1 
Solution

77. If a cos3 θ + 3α cos θ sin2 θ = m, α sin3 θ + 3α cos2 θ sin θ = n, prove that (m + n)2/3 + (m -n)2/3 
Solution

78. If x = a cos3 θ, y = b sin3 θ, prove that (x/a)2/3 θ + (y/b)2/3 = 1
Solution

79. If 3 sin θ + 5 cos θ = 5, prove that 5 sin θ - 3 cos θ = ±3. 
Solution

80. If a cos θ + b sin θ = m and a sin θ - b cos θ = n, prove that a2 + b2 = m2 + n2 
Solution

81. If cos θ + cot θ = m and cosec θ - cot θ = n, prove that m n = 1 
Solution

82. If cos A + cos2 A = 1 , prove that sin2 A + sin4 A = 1
Solution
cos A + cos2 A = 1
cos A = 1 - cos2 A
cos A = sin2 A
LHS = sin2 A + sin4 A 
= sin2 A+ sin2 A
= sin2 A + (cos A)2 
= sin2 A + cos A 
= 1 

83. Prove that : 
(i) √(sec θ - 1)/(sec θ + 1) + √(sec θ + 1)/(sec θ - 1) = 2 cosec θ
(ii) √(1 + sin θ )/(1 - sin θ) + √(1 - sin θ)/(1 + sin θ) = 2 sec θ
(iii) √(1 + cos θ )/(1 - cos θ) + √(1 - cos θ)/(1 + cos θ) = 2 cosec θ
(iv) (sec θ - 1)/(sec θ + 1) = (sinθ/(1 + cos θ))2 
Solution
(i) √(sec θ - 1)/(sec θ + 1) + √(sec θ + 1)/(sec θ - 1) = 2 cosec θ

(ii) √(1 + sin θ )/(1 - sin θ) + √(1 - sin θ)/(1 + sin θ) = 2 sec θ

(iii) √(1 + cos θ )/(1 - cos θ) + √(1 - cos θ)/(1 + cos θ) = 2 cosec θ

(iv) (sec θ - 1)/(sec θ + 1) = (sinθ/(1 + cos θ))2

84. If cos θ + cos2 θ = 1, prove that 
sin12 θ + 3sin10 θ + 3sin8 θ + sin6 θ + 2 sin4 θ + 2sin2 θ - 2 = 1 
Solution

85. Given that (1 + cos α)(1 + cos β)(1 + cosγ) = (1 - cos α)(1 - cos β)(1 - cos γ)
Show that one of the values of each member of this equality is sin α sin β sin γ
Solution

86. If sin θ + cos θ = x P.T sin6 θ + cos6 θ = [4 - 3(x2 - 1)2 /4]
Solution

87. if x = a sec θ cos ∅y = b sec θ sin∅ and z = c tan θ, S.T x2/a2 + y2/b2 - z2/c2 = 1 
Solution
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