Question

Prove that $se{c}^4\theta -ta{n}^4\theta =1+2ta{n}^2\theta .$

Answer 1

LHS

${\sec }^4\theta -{\tan }^4\theta$

$\Rightarrow ({\sec }^2\theta -{\tan }^2\theta )({\sec }^2\theta +{\tan }^2\theta )$ $[\because a^2-b^2=(a-b\left)\right(a+b\left)\right]$

$\Rightarrow 1({\sec }^2\theta +{\tan }^2\theta )$ $[\because 1+{\tan }^2\theta ={\sec }^2\theta \Rightarrow {\sec }^2\theta -{\tan }^2\theta =1]$

$\Rightarrow (1+{\tan }^2\theta +{\tan }^2\theta )$

$\Rightarrow 1+2{\tan }^2\theta =$RHS.

Hence proved.