Answer
$tan[sin^{-1}(-\frac{\sqrt 2}{2})]=-1$
Work Step by Step
Let $u=sin^{-1}(-\frac{\sqrt 2}{2})$. Then:
$sin~u=-\frac{\sqrt 2}{2}$
The range $sin^{-1}~x$ is $-\frac{\pi}{2}\lt x\lt \frac{\pi}{2}$. So, since $sin^{-1}(-\frac{\sqrt 2}{2})\lt0$, then $-\frac{\pi}{2}\lt u\lt0$ (Fourth Quadrant)
$sin^2u+cos^2u=1$
$cos^2u=1-sin^2u$
$cos^2u=1-(-\frac{\sqrt 2}{2})^2=1-\frac{2}{4}=\frac{1}{2}$
$cos~u=\frac{1}{\sqrt 2}=\frac{1}{\sqrt 2}\frac{\sqrt 2}{\sqrt 2}=\frac{\sqrt 2}{2}$ (Fourth Quadrant)
$tan[sin^{-1}(-\frac{\sqrt 2}{2})]=tan~u=\frac{sin~u}{cos~u}=\frac{-\frac{\sqrt 2}{2}}{\frac{\sqrt 2}{2}}=-1$
