Answer
$9$
Work Step by Step
As period of tangent function is $\pi$, then
$\tan \theta=\tan(\theta+\pi)=\tan(\theta+2\pi)$
Hence,
$$\tan\theta+\tan(\theta+\pi)+\tan(\theta+2\pi)=\tan{\theta}+\tan{\theta}+\tan{\theta}=3\tan\theta=3\times3=9$$
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
by Sullivan III, Michael
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
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