Answer
$csc[arctan(-\frac{5}{12})]=-\frac{13}{12}$
Work Step by Step
Let $u=arctan(-\frac{5}{12})$. Then:
$tan~u=-\frac{5}{12}$
$cot~u=\frac{1}{tan~u}=-\frac{12}{5}$
The range $arctan~x$ is $-\frac{\pi}{2}\lt x\lt\frac{\pi}{2}$. So, since $arctan(-\frac{5}{12})\lt0$, then $-\frac{\pi}{2}\lt u\lt0~~$ (Fourth Quadrant)
$csc^2u=1+cot^2u$
$csc^2u=1+(-\frac{5}{12})^2$
$csc^2u=1+\frac{25}{144}=\frac{169}{144}$
$csc~u=-\frac{13}{12}~~$ (Fourth Quadrant)
$csc[arctan(-\frac{5}{12})]=csc~u=-\frac{13}{12}$
