Answer
$-\frac{\sqrt {13}}{2}$
Work Step by Step
Given: $\tan \theta=\frac{3}{2}$
Recall: $\sec^{2}\theta=1+\tan^{2}\theta$
$\implies \sec\theta=\pm\sqrt {1+\tan^{2}\theta}$
In the quadrant III, $\cos\theta$ is negative.
So, $\sec \theta =\frac{1}{\cos\theta}$ is also negative.
$\implies \sec\theta=-\sqrt {1+\left(\frac{3}{2}\right)^{2}}$
$=-\sqrt {\frac{13}{4}}=-\frac{\sqrt {13}}{2}$
