## Precalculus (6th Edition) Blitzer

$-\dfrac{5\sqrt 3}{6}$
Solve: $\cos (\dfrac{5\pi}{6}+ 2n\pi)+\tan (\dfrac{5\pi}{6}+n\pi)$ or, $=\cos (\dfrac{5\pi}{6})+\tan (\dfrac{5\pi}{6})$ But, the reference angle is $\pi-\dfrac{5\pi}{6}=\dfrac{\pi}{6}$ So, $\cos (\dfrac{5\pi}{6})+\tan (\dfrac{5\pi}{6})=\cos (\dfrac{\pi}{6})+\tan (\dfrac{\pi}{6})$ or, $=\dfrac{-\sqrt 3}{2}+\dfrac{-\sqrt 3}{3}$ or, $=\dfrac{-\sqrt 3}{2}-\dfrac{\sqrt 3}{3}$ or, $=-\dfrac{5\sqrt 3}{6}$