## Calculus (3rd Edition)

$\cos\theta=-\frac{4}{5}$
$\csc^{2}\theta= 1+\cot^{2}\theta=1+\frac{16}{9}=\frac{25}{9}$ $\sin^{2}\theta=\frac{1}{\csc^{2}\theta}=\frac{9}{25}$ $\cos^{2}\theta=1-\sin^{2}\theta=1-\frac{9}{25}=\frac{16}{25}$ $\implies \cos\theta=±\frac{4}{5}$ 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, $\cos\theta=-\frac{4}{5}$