tan(sin^-1(-sqrt3/2))=?

$$\\tan(sin^{-1}\left(\frac{-\sqrt3}{2}\right))\\\\
\mbox{Consider }\qquad sin^{-1}\left(\frac{-\sqrt3}{2}\right)\\\\
\frac{-\pi}{2}\le sin^{-1}\left(\frac{-\sqrt3}{2}\right)\le \frac{\pi}{2}\\\\
\mbox{The inverse sine is negative so the angle is in the 4th quad}\\
\mbox{Keep this in mind but look for the equivalent 1st quadrant answer}\\$$

$$\\sin^{-1}\left(\frac{\sqrt3}{2}\right)=60^0\\\\
tan60^0=\sqrt3\\$$

But it is  not really 60degrees    $$\frac{\pi}{3}\;radians$$    because it is in the 4th quad i.e.  $$\frac{5\pi}{3}$$  radians

So therefore the answer is $$-\sqrt{3}$$

$$\\tan(sin^{-1}\left(\frac{-\sqrt3}{2}\right))=\;-\sqrt3$$

tan(asin(-sqrt(3)/2)) = -1.73205080757