Now, $\sqrt {(4{{\sin }^4}\alpha + {{\sin }^2}2\alpha )} + 4{\cos ^2}\left( {\frac{\pi }{4} - \frac{\alpha }{2}} \right)$
$ = \sqrt {(4{{\sin }^4}\alpha + 4{{\sin }^2}\alpha {{\cos }^2}\alpha )} + 2.2{\cos ^2}\left( {\frac{\pi }{4} - \frac{\alpha }{2}} \right)$
$ = \sqrt {4{{\sin }^2}\alpha ({{\sin }^2}\alpha + {{\cos }^2}\alpha )} + 2\left[ {1 + \cos \left( {\frac{\pi }{2} - \alpha } \right)} \right]$
$ = \pm 2\sin \alpha + 2 + 2\sin \alpha $
On taking $-ve$, answer is $2$ and on taking $+ve$, answer is $2 + 4\sin \alpha $
But $\pi < \alpha < \frac{{3\pi }}{4},$
Hence answer is $2 - 4\sin \alpha $ because $\sin \alpha $ is $ - ve$ in third quadrant.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.