Answer
$\frac{4\sqrt{65}}{65}$
Work Step by Step
Given: $\sec\theta=-\frac{9}{4}$
Recall: $\sec^{2}\theta -\tan^{2}\theta=1$
$\implies \tan\theta=\pm \sqrt {\sec^{2}\theta-1}$
Since both $\sin\theta$ and $\cos\theta$ is negative in the third quadrant, $\tan\theta=\frac{\sin\theta}{\cos\theta}$ is positive in this quadrant.
$\implies \tan\theta=+\sqrt {\sec^{2}\theta-1}=\sqrt {\left(-\frac{9}{4}\right)^{2}-1}$
$=\sqrt {\frac{81}{16}-1}=\sqrt {\frac{65}{16}}=\frac{\sqrt {65}}{4}$
$\cot\theta=\frac{1}{\tan\theta}=\frac{1}{\frac{\sqrt {65}}{4}}=\frac{4}{\sqrt{65}}=\frac{4\sqrt{65}}{65}$
