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

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