Answer
15a. $-\frac{\sqrt 3}{3}$
15b. $-\sqrt 3$
Work Step by Step
15a. $tan \frac {5\pi} {6}$
Reference angle is $\pi - \frac{5\pi}{6} = \frac{\pi}{6}$
Quadrant II, so tangent is negative
$-\frac{\sqrt 3}{3}$
15b. $cot \frac {5\pi} {6}$ is 1 over tangent, so the answer is $-\sqrt 3$
