Answer
$-\frac{\sqrt 3}{3}$
Work Step by Step
$\sin^{2}\theta+\cos^{2}\theta=1$
$\implies \cos^{2}\theta=1-\sin^{2}\theta$
$=1-(\frac{1}{2})^{2}=\frac{3}{4}$
$\sec^{2}\theta=\frac{1}{\cos^{2}\theta}=\frac{1}{\frac{3}{4}}=\frac{4}{3}$
$\tan^{2}\theta=\sec^{2}\theta-1=\frac{4}{3}-1=\frac{1}{3}$
$\tan\theta=\pm \frac{1}{\sqrt 3}$
$\tan\theta$ is negative in the second quadrant.
$\implies \tan\theta=-\frac{1}{\sqrt {3}}=-\frac{\sqrt 3}{3}$
