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