Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 5 - Trigonometric Functions - Section 5.3 Properties of the Trigonometric Functions - 5.3 Assess Your Understanding - Page 418: 91



Work Step by Step

As period of tangent function is $\pi$, then $\tan \theta=\tan(\theta+\pi)=\tan(\theta+2\pi)$ Hence, $$\tan\theta+\tan(\theta+\pi)+\tan(\theta+2\pi)=\tan{\theta}+\tan{\theta}+\tan{\theta}=3\tan\theta=3\times3=9$$
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