Answer
$\tan\theta=-\frac{15}{8}$
$\cot\theta=-\frac{8}{15}$
$\sec\theta=-\frac{17}{8}$
$\cos\theta=-\frac{8}{17}$
$\sin\theta=\frac{15}{17}$
$\csc\theta=\frac{17}{15}$
Work Step by Step
$\tan\theta=-\frac{15}{8}$
$\cot\theta=\frac{1}{\tan\theta}=-\frac{8}{15}$
Recall that $\tan^{2}\theta+1=\sec^{2}\theta$
$\implies\sec^{2}\theta=(-\frac{15}{8})^{2}+1=\frac{289}{64}$
$\sec\theta=\pm\sqrt {\frac{289}{64}}=-\frac{17}{8}$ ($\sec\theta\lt0$ in quadrant II)
$\cos\theta=\frac{1}{\sec\theta}=-\frac{8}{17}$
$\sin\theta=\tan\theta\cos\theta=-\frac{15}{8}\times-\frac{8}{17}=\frac{15}{17}$
$\csc\theta=\frac{1}{\sin\theta}=\frac{17}{15}$
