Question

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

• A. True
• B. False

Answer 1

The correct option is B False

${\text{sec}}^{4}A(1-{\sin }^{4}A)-2{\tan }^{2}A={\text{sec}}^{4}A(1-{\sin }^{2}A)(1+{\sin }^{2}A)-2{\tan }^{2}A$ $(\because a^2-b^2=(a-b\left)\right(a+b\left)\right)$

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

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

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

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

So the relation is $\text{False}$