Answer
$-\frac{2\sqrt {21}}{21}$
Work Step by Step
1. Let $f^{-1}(-\frac{2}{5})=t$, we have $f(t)=sin(t)=-\frac{2}{5}$.
2. Let $r=5,y=-2$, we have $x=\sqrt {5^2-(-2)^2}=\sqrt {21}$
3. Thus $h(f^{-1}(-\frac{2}{5}))=tan(t)=\frac{y}{x}=-\frac{2\sqrt {21}}{21}$
