\(\eqalign{ a)& {{2\sin 2\alpha - \sin 4\alpha } \over {2\sin 2\alpha + \sin 4\alpha }} \cr&= {{2\sin 2\alpha - 2\sin 2\alpha .cos2\alpha } \over {2\sin 2\alpha + 2\sin 2\alpha .cos2\alpha }} \cr & = {{1 - \cos 2\alpha } \over {1 + \cos 2\alpha }} = {{2{{\sin }^2}\alpha } \over {2{{\cos }^2}\alpha }} \cr&=\tan^2\alpha.\cr} \)
\(\eqalign{b)& \tan \alpha \left({{1 + {{\cos }^2}\alpha } \over {\sin \alpha }} - \sin \alpha\right ) \cr&= {{\sin \alpha } \over {\cos \alpha }}\left({{1 + {{\cos }^2}\alpha - {{\sin }^2}\alpha } \over {\sin \alpha }}\right) \cr & = {{\sin \alpha } \over {\cos \alpha }}.{{2{{\cos }^2}\alpha } \over {\sin \alpha }} = 2\cos \alpha. \cr} \)
\(d) \, {{\sin 5\alpha - \sin 3\alpha } \over {2\cos 4\alpha }} = {{2\cos {{5\alpha + 3\alpha } \over 2}\sin {{5\alpha - 3\alpha } \over 2}} \over {2\cos 4\alpha }} \)\(= \sin \alpha \)
