Trying to find the right answer to this..

(sin A + cosec A) 2 + (cos A + sec A) 2 = 7 + tan 2 A + cot 2 A L.H.S. = (sin A + cosec A) 2 + (cos A + sec A) 2 = (sin 2 A + cosec 2 A + 2 sin A. cosec A + cos 2 A + sec 2 A + 2 cos A. sec A = (sin 2 A + cos 2 A) + 2 sin A. cosec A + 2 cos A. sec A + cosec 2 A + sec 2 A = 1 + 2 + 2 + (cot 2 A + 1) + (tan 2 A + 1) = 5 + 1 + 1 + cot 2 A + tan 2 A = 7 + tan 2 A + cot 2 A = R.H.S. Hence, L.H.S. = R.H.S.

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Introduction to Trigonometry

Short Answer Type

51. Prove the following identities where the angles involved are acute angles for which the expressions are defined.
(sin A + cosec A)² + (cos A + sec A)² = 7 + tan²A + cot²A

(sin A + cosec A)² + (cos A + sec A)²= 7 + tan²A + cot² A
L.H.S.
= (sin A + cosec A)² + (cos A + sec A)²= (sin² A + cosec² A + 2 sin A. cosec A + cos² A + sec² A + 2 cos A. sec A
= (sin² A + cos² A) + 2 sin A. cosec A + 2 cos A. sec A + cosec² A + sec² A
= 1 + 2 + 2 + (cot² A + 1) + (tan² A + 1)
= 5 + 1 + 1 + cot² A + tan² A
= 7 + tan² A + cot² A = R.H.S.
Hence, L.H.S. = R.H.S.

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52. Prove the following identities where the angles involved are acute angles for which the expressions are defined.
(cosec A - sin A) (sec A - cos A) =

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Long Answer Type

53. Prove the following identities where the angles involved are acute angles for which the expressions are defined.

open parentheses fraction numerator 1 plus tan squared straight A over denominator 1 plus cot squared straight A end fraction close parentheses space equals space open parentheses fraction numerator 1 minus tanA over denominator 1 minus cotA end fraction close parentheses squared

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Short Answer Type

54. Write the value of sin² 69° + sin² 21°.

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55. If sin θ = 1/3, then find the value of (2 cot²θ+ 2).

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56. In ΔABC, ∠B = 90°
3 sin A = 3/5 , what is the value cos C.

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57. If 15 cot θ = 8, then find the value of

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58.

Simplify: $fraction numerator sin cubed straight theta plus cos cubed straight theta over denominator sinθ plus cosθ end fraction plus sinθ. space cosθ$