Question

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

Answer 1

$L.H.S.={\sec }^{2}x(se{c}^{2}x-1)$

$=se{c}^{2}x{\tan }^{2}x$

$R.H.S.=ta{n}^{2}x({\tan }^{2}x+1)$

$=ta{n}^{2}x{\sec }^{2}x$

Hence,

$L.H.S.=R.H.S.$ for all values of x