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