Answer
$\sin\theta=\frac{\sqrt {2}}{6}$
$\cos\theta=-\frac{\sqrt {34}}{6}$
$\csc\theta=3\sqrt 2$
$\sec\theta=-\frac{3\sqrt {34}}{17}$
$\tan\theta=-\frac{\sqrt {17}}{17}$
$\cot\theta=-\sqrt {17}$
Work Step by Step
$\sin\theta=\frac{\sqrt {2}}{6}$
Recall that $\sin^{2}\theta+\cos^{2}\theta=1$
$\implies\cos^{2}\theta=1-\sin^{2}\theta=1-(\frac{\sqrt 2}{6})^{2}=\frac{34}{36}$
$\cos\theta=\pm\sqrt {\frac{34}{36}}=-\frac{\sqrt {34}}{6}$ (given $\cos\theta\lt0$)
$\csc\theta=\frac{1}{\sin\theta}=\frac{6}{\sqrt {2}}=\frac{6\sqrt 2}{2}=3\sqrt 2$
$\sec\theta=\frac{1}{\cos\theta}=-\frac{6}{\sqrt {34}}=-\frac{6\sqrt {34}}{34}=-\frac{3\sqrt {34}}{17}$
$\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{\frac{\sqrt 2}{6}}{-\frac{\sqrt {34}}{6}}=-\frac{\sqrt 2}{\sqrt {34}}$
$=-\frac{\sqrt {2}\times\sqrt {34}}{34}=-\frac{\sqrt {2}\times\sqrt {2}\times\sqrt {17}}{2\times17}=-\frac{\sqrt {17}}{17}$
$\cot\theta=\frac{1}{\tan\theta}=-\frac{17}{\sqrt {17}}=-\sqrt {17}$
