$\mathrm{\theta }$ is in $3^{\mathrm{rd}}$ quadrant then $\sqrt{(4 {\sin }^4\mathrm{\theta }+{\sin }^2\mathrm{\theta })}+4{\cos }^2\left(\frac{\mathrm{\pi }}{2},\frac{\mathrm{\theta }}{2}\right)=$

$\begin{array}{rcl}\mathrm{\theta }& \in & \mathrm{Q}3\\ & & \sqrt{4 {\sin }^4\mathrm{\theta }+{\sin }^2\mathrm{\theta }}+4 {\cos }^2 \left(\frac{\mathrm{\pi }}{4}-\frac{\mathrm{\theta }}{2}\right)\\ & =& \sqrt{4{\sin }^4\mathrm{\theta }+4{\sin }^2\mathrm{\theta } {\cos }^2\mathrm{\theta }}+2\left(2{\cos }^2 \left(\frac{\mathrm{\pi }}{4}-\frac{\mathrm{\theta }}{2}\right)\right)\\ & =& -2\mathrm{sin\theta }\sqrt{{\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }}+2\left[1+\cos \left(\frac{\mathrm{\pi }}{2}-\mathrm{\theta }\right)\right]\\ & =& -2\mathrm{sin\theta }+2+2\mathrm{sin\theta }\\ & =& 2\end{array}$