## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 5 - Review Exercises - Page 708: 27

#### Answer

The expression on the left-side is equal to the expression on the right-side.

#### Work Step by Step

The given expression on the left side $\frac{\sin 2\theta }{1-{{\sin }^{2}}\theta }$ can be simplified by applying the double angle formula $\sin 2\theta =2\sin \theta \cos \theta$ and as per the Pythagorean Theorem ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$; therefore, $1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta$. So, the expression will be: \begin{align} & \frac{\sin 2\theta }{1-{{\sin }^{2}}\theta }=\frac{2\sin \theta \cos \theta }{{{\cos }^{2}}\theta } \\ & =\frac{2\sin \theta }{\cos \theta }.\frac{\cos \theta }{\cos \theta } \\ & =2\tan \theta .1 \\ & =2\tan \theta \end{align} Hence, the expression on the left-side is equal to the expression on the right-side.

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