If $\tan \theta = 3$ and 0 lies in $II{I}^{rd}$ quadrant, then the value of $\sin \theta$ is 

Given, $\tan \theta =\frac{3}{1}$ and $\theta$ lies in III quadrant. 

We know that ${\sec }^2\theta =1+{\tan }^2\theta =1+{\left(\frac{3}{1}\right)}^2=10$

$\Rightarrow \sec \theta =\pm \sqrt{10}$

Since, $\theta$ lies in III quadrant, so $sec \theta =-\sqrt{10}$

$\Rightarrow \cos \theta =\frac{1}{\sec \theta }=\frac{1}{-\sqrt{10}}$

Also, 

$$\begin{gathered} {\sin }^2\theta =1-{\cos }^2\theta =1-{\left(-\frac{1}{\sqrt{10}}\right)}^2 \\ \begin{array}{r}=1-\frac{1}{10}=\frac{9}{10}\\ \Rightarrow \sin \theta =\pm \sqrt{\frac{9}{10}}\end{array} \end{gathered}$$

Since, $\theta$ lies in III quadrant so $\sin \theta =-\sqrt{\frac{9}{10}}=\frac{-3}{\sqrt{10}}.$