Answer
$-\frac{\sqrt3}{2}$
Work Step by Step
First, we solve tan$^{-1}(-\sqrt3)$. The interval of the inverse of tangent is $-\frac{\pi}{2}\lt x\lt\frac{\pi}{2}$. Then, we must find what angle will give a slope of $-\sqrt3$ in that interval. The answer is $-\frac{\pi}{3}$
Now, we can solve sin$\left(-\frac{\pi}{3}\right)$ which is $-\frac{\sqrt3}{2}$