Question

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

Answer 1

$LHS$

$={\sec }^{2}A-\frac{{\sin }^{2}A-2{\sin }^{4}A}{2{\cos }^{4}A-{\cos }^{2}A}$

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

$={\sec }^{2}A-{\tan }^{2}A\left(\frac{\cos 2A}{\cos 2A}\right)[\because \cos 2A={\cos }^{2}A-{\sin }^{2}A]$

$={\sec }^{2}A-{\tan }^{2}A=1[\because {\sec }^{2}A-{\tan }^{2}A=1]$

$=RHS$ (proved)