## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 5 - Review Exercises - Page 707: 13

#### 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 $\frac{1-\cos t}{1+\cos t}$ can be simplified by rationalizing it. \begin{align} & \frac{1-\cos t}{1+\cos t}=\frac{1-\cos t}{1+\cos t}.\frac{1-\cos t}{1-\cos t} \\ & =\frac{{{\left( 1-\cos t \right)}^{2}}}{1-{{\cos }^{2}}t} \\ & =\frac{{{\left( 1-\cos t \right)}^{2}}}{{{\sin }^{2}}t} \\ & ={{\left( \frac{1-\cos t}{\sin t} \right)}^{2}} \end{align} Now, the expression can be simplified by applying the reciprocal identity $\frac{1}{\sin t}=\csc t$ and quotient identity $\cot t=\frac{\cos t}{\sin t}$ \begin{align} & {{\left( \frac{1-\cos t}{\sin t} \right)}^{2}}={{\left( \frac{1}{\sin t}-\frac{\cos t}{\sin t} \right)}^{2}} \\ & ={{\left( \csc t-\cot \right)}^{2}} \end{align} Hence, the expression on the left side is equal to the expression on the right side.

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