Answer
$\sin\theta$ = $\frac{1}{2}$
$\cos\theta$ = - $\frac{\sqrt 3}{2}$
$\tan\theta$ = - $\frac{1}{\sqrt 3}$
$\sec\theta$ = - $\frac{2}{\sqrt 3}$
$\cot\theta$ = - $\sqrt 3$
Work Step by Step
Given $\csc\theta$ = 2
By reciprocal identity-
$\sin\theta$ = $\frac{1}{\csc\theta}$ = $\frac{1}{2}$
We know from first Pythagorean identity that-
$\cos\theta$ = ± $\sqrt (1-\sin^{2}\theta)$
Given $\csc\theta$ is positive and $\cos\theta$ is negative, therefore $\theta$ terminates in Q II.
$\cos\theta$ = - $\sqrt (1-\sin^{2}\theta)$
substitute the 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-
$\sec\theta$ = $\frac{1}{\cos\theta}$ = $\frac{1}{-\sqrt 3/2}$ = - $\frac{2}{\sqrt 3}$
$\cot\theta$ = $\frac{1}{\tan\theta}$ = $\frac{1}{-1/\sqrt 3}$ = - $\sqrt 3$
