Question

Prove the following trigonometric identities:
(1+sinθ)²(1-sinθ)²/2cosθ=1+sin²θ/1-sin²θ

Answer 1

we have to prove that,

(1 + sinθ)²(1 - sinθ)²/2cosθ = (1 + sin²θ)/(1 - sin²θ)

LHS = (1 + sinθ)²(1 - sinθ)/2cosθ

= {(1 + sinθ)(1 - sinθ)}²/2cosθ

= {1 - sin²θ}²/2cosθ

we know, sin²x + cos²x = 1 so, 1 - sin²θ = cos²θ

= {cos²θ}²/2cosθ

= cos³θ/2

RHS = (1 + sin²θ)/(1 - sin²θ)

= (1 + 1 + cos²θ)/cos²θ

= (2 + cos²θ)/cos²θ

= 2sec²θ + 1

LHS ≠ RHS

hence, (1 + sinθ)²(1 - sinθ)²/2cosθ ≠ (1 + sin²θ)/(1 - sin²θ)

you can check it putting value of θ

let put θ = 30°

LHS = (1 + 1/2²)(1 - 1/2²)/2 × √3/2 = (15/16)/√3 = 5√3/16

RHS = (1 + 1/2²)/(1 - 1/2²) = 5/3

here also LHS ≠ RHS

Answer 2

Answer:

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