Question

${\sec }^6\theta -{\tan }^6\theta =1+3{\sec }^2\theta {\tan }^2\theta$

• A. True
• B. False

Answer 1

The correct option is A True

As we know that

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

${\text{sec}}^6\theta -{\tan }^6\theta =({\text{sec}}^2\theta {)}^3-({\tan }^2\theta {)}^3$

$=({\text{sec}}^2\theta -{\tan }^2\theta )({\text{sec}}^4\theta +{\tan }^4\theta +{\text{sec}}^2\theta {\tan }^2\theta )$

$=({\text{sec}}^2\theta -{\tan }^2\theta {)}^2+2{\sec }^2\theta {\tan }^2\theta +{\text{sec}}^2\theta {\tan }^2\theta$

$=1+3{\text{sec}}^2\theta {\tan }^2\theta$

So the given relation is $\text{True}$