Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 6 - Trigonometric Functions - 6.3 Properties of the Trigonometric Functions - 6.3 Assess Your Understanding - Page 395: 91



Work Step by Step

We know that $\tan$ has a period of $\pi$, hence $\tan(\theta)=\tan(\theta+k\pi)$ where $k$ is an integer. Therefore, $\tan(\theta)+\tan(\theta+\pi)+\tan(\theta+2\pi)\\ =\tan(\theta)+\tan(\theta)+\tan(\theta)\\ =3+3+3\\ =9$
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