Answer
$\tan\theta$ = $\frac{\sqrt {1-\cos^{2}\theta}}{\cos\theta}$
($\sin\theta$ and $\cos\theta$ both will be negative making $\tan\theta$ positive and $\cos\theta\ne0$ as $\theta$ lies in quadrant III)
Work Step by Step
We are supposed to write $\tan\theta$ in terms of $\cos\theta$ while $\theta$ lies in Quadrant III.
Using ratio identity for $\tan$-
$\tan\theta$ = $\frac{\sin\theta}{\cos\theta}$
($\cos\theta\ne0$ as $\theta$ lies in quadrant III)
From Pythagorean identity-
$\sin\theta$ may be written as $\sqrt {1-\cos^{2}\theta}$
Therefore-
$\tan\theta$ = $\frac{\sqrt {1-\cos^{2}\theta}}{\cos\theta}$
($\sin\theta$ and $\cos\theta$ both will be negative making $\tan\theta$ positive and $\cos\theta\ne0$ as $\theta$ lies in quadrant III)
