If $ heta$ is in the third quadrant, then $\sqrt{4 \sin ^4 heta+\sin ^2 2 heta}+4 \cos ^2\left(\frac{\pi}{4}-\frac{ heta}{2} ight)=$

Question

If $	heta$ is in the third quadrant, then $\sqrt{4 \sin ^4 	heta+\sin ^2 2 	heta}+4 \cos ^2\left(\frac{\pi}{4}-\frac{	heta}{2}ight)=$

Aicharya · Accepted Answer

No solution. Refer to answer key.

If $\theta$ is in the third quadrant, then $\sqrt{4 \sin ^4 \theta+\sin ^2 2 \theta}+4 \cos…