Chapter 5, Problem 128CR

To determine

To evaluate: if $\sec \theta =\sqrt{1+{\tan }^2\theta }$

is an identity.

Answer to Problem 128CR

Yes, it is identity.

Explanation of Solution

Given:

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

It is given that

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

Taking right hand side,

  $\begin{array}{c}\sqrt{1+{\tan }^2\theta }=\sqrt{1+\frac{{\sin }^2\theta }{{\cos }^2\theta }}\\ \sqrt{1+\frac{{\sin }^2\theta }{{\cos }^2\theta }}=\sqrt{\frac{{\cos }^2\theta +{\sin }^2\theta }{{\cos }^2\theta }}\\ \sqrt{\frac{{\cos }^2\theta +{\sin }^2\theta }{{\cos }^2\theta }}=\sqrt{\frac{1}{{\cos }^2\theta }}\\ \sqrt{\frac{1}{{\cos }^2\theta }}=\sqrt{{\sec }^2\theta }\\ \sqrt{{\sec }^2\theta }=\sec \theta \end{array}$

So,

Yes, it is identity.