Question

Prove: ${\sec }^4\theta -{\tan }^4\theta =2{\sec }^2\theta -1$

Answer 1

$LHS=({\sec }^4\theta -{\tan }^4\theta )$

$=({\sec }^2\theta {)}^2-({\tan }^2\theta {)}^2$

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

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

$=2{\sec }^2\theta -1$

$=RHS$