## University Calculus: Early Transcendentals (3rd Edition)

$-\frac{\sqrt 3}{3}$
The given problem can be solved as: $tan({sin^{-1}(-\frac{1}{2})})=tan ( -\frac{\pi}{6})$ or, $tan ( -\frac{\pi}{6})=\frac{-\sqrt 3}{3}$ Hence. the final answer is: $-\frac{\sqrt 3}{3}$