Question

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

Answer 1

Consider the L.H.S,

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

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

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

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

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

$=[\left({\sec }^{2}A\right)-\left({\tan }^{2}A\right)][\left({\sec }^{2}A\right)+\left({\tan }^{2}A\right)]-2{\tan }^{2}A$

We know that

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

Therefore,

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

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

$=1$

Hence, proved