Question

Prove:

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

Answer 1

According to problem,

$\begin{array}{c}{\sec }^6\theta \\ ={\left({\sec }^2\theta \right)}^3\\ ={\left(1+{\tan }^2\theta \right)}^3\\ ={\left(1\right)}^3+{\left({\tan }^2\theta \right)}^3+3\times 1\times {\tan }^2\theta \left(1+{\tan }^2\theta \right)\\ =1+{\tan }^6\theta +3{\tan }^2\theta {\sec }^2\theta \end{array}$

Hence proved