Answer
$$\sin\theta=-\frac{\sqrt{15}}{5}$$
Work Step by Step
$$\tan\theta=-\frac{\sqrt6}{2}\hspace{1cm}\cos\theta\gt0$$
From a Pythagorean Identity
$$\tan^2\theta+1=\sec^2\theta$$
combined with a Reciprocal Identity
$$\sec\theta=\frac{1}{\cos\theta}$$
we have
$$\tan^2\theta+1=\frac{1}{\cos^2\theta}$$
$$\frac{1}{\cos^2\theta}=(-\frac{\sqrt 6}{2})^2+1=\frac{6}{4}+1=\frac{10}{4}$$
$$\cos^2\theta=\frac{4}{10}$$
$$\cos\theta=\frac{2}{\sqrt{10}}=\frac{2\sqrt{10}}{10}=\frac{\sqrt{10}}{5}$$
Also, from Quotient Identities,
$$\tan\theta=\frac{\sin\theta}{\cos\theta}$$
which means
$$\sin\theta=\tan\theta\times\cos\theta$$
$$\sin\theta=(-\frac{\sqrt6}{2})\times(\frac{\sqrt{10}}{5})=-\frac{\sqrt{60}}{10}=-\frac{2\sqrt{15}}{10}=-\frac{\sqrt{15}}{5}$$
