Answer
$\sec\theta$ = - $\frac{25}{24}$
Work Step by Step
We know from third Pythagorean identity that-
$\sec\theta$ = ± $\sqrt (1+\tan^{2}\theta)$
As $\cos\theta\lt0$, i.e. $\cos\theta$ is negative and $\tan\theta$ is positive, hence $\theta$ terminates in Q III, Therefore $\sec\theta$ will be negative-
$\sec\theta$ = - $\sqrt (1+\tan^{2}\theta)$
substitute the given value of $\tan\theta$-
$\sec\theta$ = - $\sqrt (1+(\frac{7}{24})^{2})$
$\sec\theta$= - $\sqrt (1+\frac{49}{576})$
$\sec\theta$ = - $\sqrt (\frac{576+49}{576})$ = $\sqrt (\frac{625}{576})$
$\sec\theta$ = - $\frac{25}{24}$