Answer
$\csc \theta=2$
$\cos \theta= -\frac{\sqrt 3}{2}$
$\sec \theta=-\frac{2}{\sqrt 3}$
$\tan \theta=-\frac{1}{\sqrt 3}$
$\cot\theta=-\sqrt 3$
Work Step by Step
Since $\sin \theta=\frac{1}{2}$, we have $\csc \theta=\frac{1}{\sin \theta}=2$
Now $\sin^{2}\theta+\cos^{2}\theta=1$, i.e., $\cos^{2}\theta=1-\sin^{2}\theta$
or $\cos^{2}\theta= 1-\frac{1}{4}=\frac{3}{4}$
Since $\theta$ terminates in QII, $\cos\theta$ is negative.
$\cos \theta= -\frac{\sqrt 3}{2}$
$\sec \theta=\frac{1}{\cos\theta}=-\frac{2}{\sqrt 3}$
Further, we have
$\tan \theta=\frac{\sin\theta}{\cos\theta}=\frac{\frac{1}{2}}{-\frac{\sqrt 3}{2}}=-\frac{1}{\sqrt 3}$ and $\cot\theta=\frac{1}{\tan\theta}=-\sqrt 3$