Question

Prove that $\to {\sec }^2\theta +{\cot }^2(90-\theta )=2{\csc }^2(90-\theta )-1$

Answer 1

LHS

$={\sec }^2\theta +{\cot }^2(90-\theta )$

$={\sec }^2\theta +{\tan }^2\theta$

RHS

$=2\cos e{c}^2(90-\theta )-1=2{\sec }^2\theta -1$

$=2\left(1+{\tan }^2\theta \right)-1=2{\tan }^2\theta +1$

$={\tan }^2\theta +1+{\tan }^2\theta ={\sec }^2\theta +{\tan }^2\theta =LHS$

Hence prove.