## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 5 - Section 5.3 - Double-Angle, Power-Reducing, and Half-Angle Formulas - Exercise Set - Page 680: 65

#### Answer

See the explanation below.

#### Work Step by Step

Let us consider the right side of the given expression: $\frac{\sin x}{1-\cos x}$ By using the trigonometric identity $\tan \frac{x}{2}=\frac{1-\cos x}{\sin x}$, the above expression can be further simplified by multiplying the numerator and denominator by $\frac{1}{\sin x}$ \begin{align} & \frac{\sin x}{1-\cos x}=\frac{\left( \sin x \right)\times \frac{1}{\sin x}}{\left( 1-\cos x \right)\times \frac{1}{\sin x}} \\ & =\frac{\frac{\sin x}{\sin x}}{\frac{1-\cos x}{\sin x}} \\ & =\frac{1}{\tan \frac{x}{2}} \\ & =\cot \frac{x}{2} \end{align} Hence, the left side of the given expression is equal to the right side, which is $\cot \frac{x}{2}=\frac{\sin x}{1-\cos x}$.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.