# How do you evaluate  tan(sin^-1(-1/6))?

$- \frac{1}{\sqrt{35}}$.
Let $a = {\sin}^{- 1} \left(- \frac{1}{6}\right)$. Then, $\sin a = - \frac{1}{6} < 0$. $a$ is in the 3rd quadrant or in the 4th. On the other hand, he "principal branch" of the inverse sine corresponds to an angle in the first or fourth quadrant, not the third. So we pick the fourth quadrant angle, and $\cos a = + \frac{\sqrt{35}}{6}$.
The given expression $= \tan a = \sin \frac{a}{\cos} a = - \frac{1}{\sqrt{35}}$.