Question

Prove ${\sec }^{4}x-{\sec }^{2}x={\tan }^{4}x+{\tan }^{2}x$

Answer 1

As we know that

${\sec }^{2}x-{\tan }^{2}x=1$,

Or

${\sec }^{2}x=1+{\tan }^{2}x$, ......(1)

Or

${\sec }^{2}x-1={\tan }^{2}x$ ......(2)

Given that:

${\sec }^{4}x-{\sec }^{2}x={\tan }^{4}x+{\tan }^{2}x$

$LHS={\sec }^{4}x-{\sec }^{2}x$

$={\sec }^{2}x({\sec }^{2}x-1)$

$=\left({\sec }^{2}x\right)\left({\tan }^{2}x\right)$ [From equation (2)]

$RHS={\tan }^{4}x+{\tan }^{2}x$

$={\tan }^{2}x({\tan }^{2}x+1)$

$=\left({\tan }^{2}x\right)\left({\sec }^{2}x\right)$ [From equation (1)]

$=\left({\sec }^{2}x\right)\left({\tan }^{2}x\right)$

Hence, $LHS=RHS$.