Answer
$\sin\theta$ = - $\frac{2\sqrt 2}{3}$
$\tan\theta$ = $2\sqrt 2$
$\csc\theta$ = -$\frac{3}{2\sqrt2}$
$\sec\theta$ = -3
$\cot\theta$ = $\frac{1}{2\sqrt 2}$
Work Step by Step
We know from first Pythagorean identity that-
$\sin\theta$ = ± $\sqrt (1-\cos^{2}\theta)$
Given $\cos\theta$ is negative and $\theta$ is not in QII, therefore $\theta$ terminates in Q III. Hence $\sin\theta$ will be negative-
OR
$\sin\theta$ = - $\sqrt (1-\cos^{2}\theta)$
substitute the given value of $\cos\theta$-
$\sin\theta$= - $\sqrt (1-(\frac{-1}{3})^{2})$
$\sin\theta$ = - $\sqrt (1-\frac{1}{9})$
$\sin\theta$ = - $\sqrt (\frac{9-1}{9})$ = - $\sqrt (\frac{8}{9})$
$\sin\theta$ = - $\frac{2\sqrt 2}{3}$
By ratio identity-
$\tan\theta$ = $\frac{\sin\theta}{\cos\theta}$
Substituting the values of $\sin\theta$ and $\cos\theta$-
$\tan\theta$ = $\frac{-2\sqrt 2/3}{ -1/3}$ = $2\sqrt 2$
From reciprocal identities-
$\csc\theta$ = $\frac{1}{\sin\theta}$ = $\frac{1}{-2\sqrt2/3}$ = -$\frac{3}{2\sqrt2}$
$\sec\theta$ = $\frac{1}{\cos\theta}$ = $\frac{1}{-1/3}$ = -3
$\cot\theta$ = $\frac{1}{\tan\theta}$ = $\frac{1}{2\sqrt 2}$
