Question

tan theta + 1 by cos theta whole square + tan theta - 1 by cos theta whole square equals to 2 of 1 + sin square theta by 1 minus sin square theta​

Answer 1

Given:

tan theta + 1 by cos theta whole square + tan theta - 1 by cos theta whole square equals to 2 of 1 + sin square theta by 1 minus sin square theta​

To find:

tan theta + 1 by cos theta whole square + tan theta - 1 by cos theta whole square equals to 2 of 1 + sin square theta by 1 minus sin square theta​

Solution:

From given, we have,

tan theta + 1 by cos theta whole square + tan theta - 1 by cos theta whole square equals to 2 of 1 + sin square theta by 1 minus sin square theta​

⇒ (tanθ + 1/cosθ)² + (tanθ - 1/cosθ)² = 2 [(1+sin²θ)/(1-sin²θ)]

Now consider, LHS,

= (tanθ + 1/cosθ)² + (tanθ - 1/cosθ)²

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

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

= [(1+sinθ)² + (1-sinθ)²]/(cosθ)²

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

= (2+2sin²θ)/(cosθ)²

= 2(1+sin²θ)/cos²θ

= 2(1+sin²θ)/(1-sin²θ)

= 2 [(1+sin²θ)/(1-sin²θ)]

= RHS

Hence the proof.