If $\cos \theta=-\frac{1}{2}$ and $\theta$ lies in third quadrant, then what will be the value of $\sin \theta+\tan \theta$

Here, $\cos \theta=-\frac{1}{2}$ and $\theta$, lies in third quadrant

Consider ΔABC, Using Pythagoras theorem,

$\mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}$

$2^{2}=(1)^{2}+\mathrm{BC}^{2}$

$\Rightarrow \mathrm{BC}^{2}=4-1$

$\mathrm{BC}^{2}=3$

$\mathrm{BC}=\sqrt{3}$

$\tan \theta+\sin \theta=-\sqrt{3}+\left\{-\left(\frac{\sqrt{3}}{2}\right)\right\}$

In third quadrant sinθ is negative and tanθ is positive.

$=\frac{\sqrt{3}}{2}$