## Algebra 2 (1st Edition)

$-\sqrt 3$
Here, we have $\theta=\dfrac{8 \pi}{3}$ Here, $\theta'=\dfrac{8 \pi}{3}-2 \pi=\dfrac{2 \pi}{3}$ and $\theta''=\pi-\theta'=\pi-\dfrac{2 \pi}{3}=\dfrac{\pi}{3}$ Now, $\cos \theta'=\cos 30^{\circ}=\dfrac{\sqrt 3}{2}$ The function tan is negative in Quadrant $II$. Thus, $\tan \theta= -\tan \theta'= -\tan \theta''=-\tan \dfrac{\pi}{3}=-\sqrt 3$