Chapter 13.6, Problem 32WE

To determine

To Prove: The given identity.

Explanation of Solution

Given:

The given identity is ${\sin }^2\alpha \left({\csc }^2\alpha +{\sec }^2\alpha \right)={\sec }^2\alpha$ .

Calculation:

Consider the given identity is,

  ${\sin }^2\alpha \left({\csc }^2\alpha +{\sec }^2\alpha \right)={\sec }^2\alpha$

Solve the above equation as,

  $\begin{array}{l}{\sin }^2\alpha \left({\csc }^2\alpha +{\sec }^2\alpha \right)={\sec }^2\alpha \\ {\sin }^2\alpha {\csc }^2\alpha +{\sin }^2\alpha {\sec }^2\alpha ={\sec }^2\alpha \\ {\sin }^2\alpha \frac{1}{{\sin }^2\alpha }+{\sin }^2\alpha \frac{1}{{\cos }^2\alpha }={\sec }^2\alpha \\ 1+\frac{{\sin }^2\alpha }{{\cos }^2\alpha }={\sec }^2\alpha \end{array}$

By the trigonometric identity the given equation is,

  $\begin{array}{l}\frac{{\cos }^2\alpha +{\sin }^2\alpha }{{\cos }^2\alpha }={\sec }^2\alpha \\ \frac{1}{{\cos }^2\alpha }={\sec }^2\alpha \\ {\sec }^2\alpha ={\sec }^2\alpha \end{array}$

Hence, the given identity is proved.