Answer
$\cos\theta$ = $\frac{\sqrt 3}{2}$
$\tan\theta$ = - $\frac{1}{\sqrt 3}$
$\csc\theta$ = -2
$\sec\theta$ = $\frac{2}{\sqrt 3}$
$\cot\theta$ = - $\sqrt 3$
Work Step by Step
We know from first Pythagorean identity that-
$\cos\theta$ = ± $\sqrt (1-\sin^{2}\theta)$
Given $\sin\theta$ is negative and $\theta$ is not in QIII, therefore $\theta$ terminates in Q IV. Hence $\cos\theta$ will be positive-
OR
$\cos\theta$ = $\sqrt (1-\sin^{2}\theta)$
substitute the given value of $\sin\theta$-
$\cos\theta$ = $\sqrt (1-(\frac{-1}{2})^{2})$
$\cos\theta$ = $\sqrt (1-\frac{1}{4})$
$\cos\theta$ = $\sqrt (\frac{4-1}{4})$ = $\sqrt (\frac{3}{4})$
$\cos\theta$ = $\frac{\sqrt 3}{2}$
By ratio identity-
$\tan\theta$ = $\frac{\sin\theta}{\cos\theta}$
Substituting the values of $\sin\theta$ and $\cos\theta$-
$\tan\theta$ = $\frac{-1/2}{\sqrt 3/2}$ = - $\frac{1}{\sqrt 3}$
From reciprocal identities-
$\csc\theta$ = $\frac{1}{\sin\theta}$ = $\frac{1}{-1/2}$ = -2
$\sec\theta$ = $\frac{1}{\cos\theta}$ = $\frac{1}{\sqrt 3/2}$ = $\frac{2}{\sqrt 3}$
$\cot\theta$ = $\frac{1}{\tan\theta}$ = $\frac{1}{-1/\sqrt 3}$ = - $\sqrt 3$
