Answer
$\sec \theta = -\frac{4}{3}$
Work Step by Step
Given $\tan\theta = \frac{\sqrt 7}{3}$ and $\theta $ is in quadrant III.
Using Identity,
$\sec^{2}\theta = 1 + \tan^{2}\theta$
$\sec^{2}\theta = 1 + (\frac{\sqrt 7}{3})^{2}$
$ = 1 + (\frac{ 7}{9})$
$= \frac{9+7}{9} = \frac{16}{9}$
$\sec^{2}\theta = \frac{16}{9}$
$\sec \theta = +\frac{4}{3} $ or $-\frac{4}{3} $
In quadrant III $\sec \theta $ is negative. Therefore,
$\sec \theta =-\frac{4}{3} $
