Answer
$tan[arccos(-\frac{2}{3})]=tan~u=-\frac{\sqrt 5}{2}$
Work Step by Step
Let $u=arccos(-\frac{2}{3})$. Then:
$cos~u=-\frac{2}{3}$
The range of $arccos~x$ is $0\leq x\leq \pi$. So, since $arccos(-\frac{2}{3})\gt\frac{\pi}{2}$, then $\frac{\pi}{2}\lt u\lt \pi$ (Second Quadrant)
$cos^2u+sin^2u=1$
$sin^2u=1-cos^2u$
$sin^2u=1-cos^2u$
$sin^2u=1-(-\frac{2}{3})^2=1-\frac{4}{9}=\frac{5}{9}$
$sin~u=\frac{\sqrt 5}{3}~~$ (Second Quadrant)
$tan[arccos(-\frac{2}{3})]=tan~u=\frac{sin~u}{cos~u}=\frac{\frac{\sqrt 5}{3}}{-\frac{2}{3}}=-\frac{\sqrt 5}{2}$
