Answer
$-\frac{3\sqrt {10}}{10}$
Work Step by Step
Step 1. Letting $tan^{-1}(-3)=t$, we have $tan(t)=-3$, and $t$ is in quadrant IV.
Step 2. Let $x=1, y=3$, and $r=\sqrt {x^2+y^2}=\sqrt {10}$.
Step 3. Sides $x,y,r$ form a right triangle with angle $|t|$ facing $y$.
Step 4. We know $sin(t)$ is negative and we have $sin(t)=-\frac{y}{r}=-\frac{3\sqrt {10}}{10}$ or $sin(tan^{-1}(-3))=-\frac{3\sqrt {10}}{10}$
