Answer
$\csc\theta$ = $-\sqrt {1+\cot^{2}\theta}$
($\csc\theta\lt0$ as $\theta$ lies in quadrant III)
Work Step by Step
We are supposed to write $\csc\theta$ in terms of $\cot\theta$ while $\theta$ lies in Quadrant III.
From Pythagorean identity, we know that-
$1 + \cot^{2}\theta = \csc^{2}\theta$
Therefore-
$\csc\theta$ = $\sqrt {1+\cot^{2}\theta}$
($\csc\theta\lt0$ as $\theta$ lies in quadrant III)