Question

Prove that
${\sec }^2\theta -\frac{{\sin }^2\theta -2\sin 4\theta }{2\cos 4\theta -{\cos }^2\theta }=1$

Answer 1

$\frac{{\sin }^2\theta -2\sin 4\theta }{2\cos 4\theta -{\cos }^2\theta }$

on simplifying using $\sin 4\theta ,\cos 4\theta$ formulae in terms of ${\sin }^2\theta ,{\cos }^2\theta$ gives

$=\frac{{\sin }^2\theta }{{\cos }^2\theta }={\tan }^2\theta$

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

$\therefore$ Hence proved