$\cos \theta=\frac{-3}{5}$
$\begin{aligned} & \pi < \theta < \frac{3 \pi}{2} \\ & \frac{\pi}{2} < \frac{\theta}{2} < \frac{3 \pi}{4} \text { i.e. } 2^{\text {nd }} \text { quadrant } \\ & \sin \frac{\theta}{2}>0, \cos \frac{\theta}{2} < 0, \tan \frac{\theta}{2} < 0 \\ & \cos \theta=1-2 \sin ^2 \frac{\theta}{2} \Rightarrow \frac{-3}{5}=1-2 \sin ^2 \frac{\theta}{2} \\ & \sin \frac{\theta}{2}=\frac{2}{\sqrt{5}} \Rightarrow \cos \theta=2 \cos ^2 \frac{\theta}{2}-1 \\ & \frac{-3}{5}=2 \cos ^2 \frac{\theta}{2}-1 \Rightarrow \cos \frac{\theta}{2}=-\frac{1}{\sqrt{5}} \\ & \tan \frac{\theta}{2}=\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}=-2 \quad \therefore \tan \frac{\theta}{2}+\sin \frac{\theta}{2}+2 \cos \frac{\theta}{2} \\ & =-2+\frac{2}{\sqrt{5}}-\frac{2}{\sqrt{5}}=-2\end{aligned}$