Answer
$\cos\theta=-\frac{4}{5}$
Work Step by Step
$\csc^{2}\theta= 1+\cot^{2}\theta=1+\frac{16}{9}=\frac{25}{9}$
$\sin^{2}\theta=\frac{1}{\csc^{2}\theta}=\frac{9}{25}$
$\cos^{2}\theta=1-\sin^{2}\theta=1-\frac{9}{25}=\frac{16}{25}$
$\implies \cos\theta=±\frac{4}{5}$
As $\cot \theta=\frac{4}{3}$, $\cot\theta>0$
$\implies \frac{\cos\theta}{\sin\theta}>0$
But $\sin\theta<0$ (given)
This implies that $\cos\theta$ is also less than 0.
Therefore, $\cos\theta=-\frac{4}{5}$
