Question

Prove that ${\sec }^{4}A(1-{\sin }^{4}A)-2{\tan }^{2}A=1$

Answer 1

L.H.S$={\sec }^{4}A(1-{\sin }^{4}A)-2{\tan }^{2}A$

We know that,

$a^2-b^2=(a+b)(a-b)$

Therefore,

$={\sec }^{4}A(1-{\sin }^{2}A)(1+{\sin }^{2}A)-2{\tan }^{2}A$

$=\frac{{\sec }^{2}A}{{\cos }^{2}A}\left({\cos }^{2}A\right)(1+{\sin }^{2}A)-2{\tan }^{2}A[\because 1-{\sin }^{2}x={\cos }^{2}x]$

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

$={\sec }^{2}A+{\tan }^{2}A-2{\tan }^{2}A$

$={\sec }^{2}A-{\tan }^{2}A$

$=1$

$=RHS$

Hence, it proved.