## Precalculus (10th Edition)

$0$
We know that $tan$ has a period of $\pi$, hence first we try and find a value where the argument is between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$. Therefore $tan(6\pi)=tan(6\pi-2\pi)=tan(4\pi)=tan(4\pi-2\pi)=tan(2\pi)=tan(2\pi-2\pi)=tan(0)=0.$