Chapter 13 - The Trigonometric Functions - 13.1 Definitions of the Trigonometric Functions - 13.1 Exercises - Page 678: 46

$\tan \frac{-5\pi}{6}=\frac{\sqrt 3}{3}$

We are given $\tan \frac{-5\pi}{6}$ $\cot(\frac{-5\pi}{6}+\pi)=\cot(\frac{\pi}{6})$ convert $\frac{\pi}{4}$ radians to degrees $\frac{\pi}{6}=\frac{\pi}{6}(\frac{180^\circ}{\pi})$ $\frac{\pi}{6}=30^\circ$ $\tan \frac{-5\pi}{6}=\tan 30^\circ$ $\tan30^\circ=\frac{\sqrt 3}{3}$ so $\tan \frac{-5\pi}{6}=\frac{\sqrt 3}{3}$