• # question_answer $2\,{{\sin }^{2}}\beta +4\,\,\cos \,(\alpha +\beta )\,\,\sin \,\alpha \,\sin \,\beta +\cos \,2\,(\alpha +\beta )=$ [MNR 1993; IIT 1977] A) $\sin \,\,2\alpha$ B) $\cos \,\,2\beta$ C) $\cos \,\,2\alpha$ D) $\sin \,\,2\beta$

$\cos 2(\alpha +\beta )=2{{\cos }^{2}}(\alpha +\beta )-1,\,2{{\sin }^{2}}\beta =1-\cos 2\beta$ L.H.S. $=-\cos 2\beta +2\cos (\alpha +\beta )\,[2\sin \alpha \sin \beta +\cos (\alpha +\beta )]$ $=-\cos 2\beta +2\cos (\alpha +\beta )\cos (\alpha -\beta )$ $=-\cos 2\beta +(\cos 2\alpha +\cos 2\beta )=\cos 2\alpha$.