## Precalculus (6th Edition) Blitzer

$\frac{2\sqrt 3}{3}$
($\frac{\sqrt 3}{2}$ , -$\frac{1}{2}$) sec $\frac{11\pi}{6}$ is computed by $\frac{1}{x}$. So sec $\frac{11\pi}{6}$ = $\frac{1}{\frac{\sqrt 3}{2}}$ = $\frac{2\sqrt 3}{3}$.