Answer
$-2$
Work Step by Step
As $\frac{8\pi}{3}=2\pi+\frac{2\pi}{3}$ and $\frac{2\pi}{3}$ is in quadrant II with a reference angle $\pi-\frac{2\pi}{3}=\frac{\pi}{3}$, we have $sec(\frac{8\pi}{3})=sec(\frac{2\pi}{3})=\frac{1}{cos(\frac{2\pi}{3})}=\frac{1}{-cos(\frac{\pi}{3})}=-2$
