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
Given $\sin\theta$ = $\frac{1}{2}$
From first Pythagorean identity-
$\cos\theta$ = ± $\sqrt {1-\sin^{2}\theta}$
As $\theta$ terminates in QII, therefore $\cos\theta$ is negative-
$\cos\theta$ = - $\sqrt {1-\sin^{2}\theta}$
= - $\sqrt {1-(\frac{1}{2})^{2}}$
= - $\sqrt {1-\frac{1}{4}}$
= - $\sqrt {\frac{4-1}{4}}$
= - $\sqrt {\frac{3}{4}}$
$\cos\theta$ = - $\frac{\sqrt 3}{2}$
From ratio identity-
$\tan\theta$ = $\frac{\sin\theta}{\cos\theta}$
= $\frac{1/2}{-\sqrt 3/2}$
$\tan\theta$ = -$\frac{1}{\sqrt 3}$
Using reciprocal identities-
$\csc\theta$ = $\frac{1}{\sin\theta}$
$\csc\theta$ = $\frac{1}{1/2}$ = 2
$\sec\theta$ = $\frac{1}{\cos\theta}$
$\sec\theta$ = $\frac{1}{-\sqrt 3/2}$ = -$\frac{2}{\sqrt 3}$
$\cot\theta$ = $\frac{1}{\tan\theta}$
$\cot\theta$ = $\frac{1}{-1/\sqrt 3}$ = - $\sqrt 3$
