Answer
$\dfrac{-3}{4}$
Work Step by Step
We know that
$ f(x)=\sin(x), h(x)=\tan(x)$
Thus we can write:
$h(f^{-1} (\dfrac{ -3}{5}))=\tan (\sin^{-1}(\dfrac{ -3}{5})) $
Suppose that $-\theta=\sin^{-1} (\dfrac{-3}{5})$
We know from SOH-CAH-TOA:
$\sin{\theta}=\dfrac{opposite}{hypotenuse}=\dfrac{3}{5}$
To find the missing adjacent side, we use the Pythagorean Theorem:
$adjacent^2+3^2=5^2$
$adjacent=\sqrt{(5)^2-(3)^2}=\sqrt {16}=4$
Therefore,
$\tan (-\theta)=-\dfrac{opposite}{adjacent}=\dfrac{-3}{4}$
Note that we are in quadrant IV because sine is negative, so the tangent of the angle should be negative, as expected.
