Question

Prove the following trigonometric identities. $\frac{(1+sin\Theta)^{2}+(1-sin\Theta)^{2} }{2cos^{2}\Theta }=\frac{1+sin^{2}\Theta }{1-sin^{2}\Theta }$

Answer 1

Answer with Step-by-step explanation:

Given : (1 + sinθ)² +(1− sinθ)² /(2cos²θ) = (1 + sin²θ)/(1−sin²θ)

LHS = (1 + sinθ)² +(1− sinθ)² /(2cos²θ)

= [(1 + 2sinθ + sin²θ) + (1 − 2sinθ + sin²θ)] / 2cos²θ

[By using identity , (a + b)² = a² + 2ab + b²]

= [1 + 2sinθ + sin²θ + 1 − 2sinθ + sin²θ ] / 2cos²θ

= [1 + 1 + 2sinθ  − 2sinθ + sin²θ + sin²θ ] / 2cos²θ

= 2 + 2sin²θ / 2cos²θ

= 2(1 + sin²θ)/ 2(1− sin²θ)

[By using the identity, cos²θ = (1- sin²θ)]

= (1+ sin²θ) / (1− sin²θ)

(1 + sinθ)² +(1− sinθ)² /(2cos²θ) = (1 + sin²θ)/(1−sin²θ)

L.H.S = R.H.S  

Hence Proved..

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

Answer:

Refer to the attachment for your answer.