## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 5 - Section 5.2 - Sum and Difference Formulas - Exercise Set - Page 669: 32

#### Answer

The expression $\frac{\tan \frac{\pi }{5}+\tan \frac{4\pi }{5}}{1-\tan \frac{\pi }{5}\tan \frac{4\pi }{5}}$ is written as $\tan \pi$ and the exact value of $\tan \pi$ is $0$.

#### Work Step by Step

Use the sum formula of tangent and rewrite the expression as the sum of angles to obtain the tangent of the angle as, \begin{align} & \tan \left( \frac{\pi }{5}+\frac{4\pi }{5} \right)=\frac{\tan \frac{\pi }{5}+\tan \frac{4\pi }{5}}{1-\tan \frac{\pi }{5}\tan \frac{4\pi }{5}} \\ & \tan \left( \pi \right)=\frac{\tan \frac{\pi }{5}+\tan \frac{4\pi }{5}}{1-\tan \frac{\pi }{5}\tan \frac{4\pi }{5}} \end{align} Therefore, the expression $\frac{\tan \frac{\pi }{5}+\tan \frac{4\pi }{5}}{1-\tan \frac{\pi }{5}\tan \frac{4\pi }{5}}$ is equivalent to $\tan \pi$. From the knowledge of trigonometric ratios defined for tangent of an angle, the exact value of $\tan \pi$ is $0$.

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