Question

Prove that (or prove the identity) ${\sec }^{4}A-{\sec }^{2}A={\tan }^{4}A+{\tan }^{2}A$.

Answer 1

Consider LHS

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

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

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

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

$={\tan }^{4}A+{\tan }^{2}A$

$=RHS$

Hence proved