A) \[4{{\cos }^{2}}\frac{\alpha -\beta }{2}\]
B) \[4{{\sin }^{2}}\frac{\alpha -\beta }{2}\]
C) \[4{{\cos }^{2}}\frac{\alpha +\beta }{2}\]
D) \[4{{\sin }^{2}}\frac{\alpha +\beta }{2}\]
Correct Answer: A
Solution :
\[{{(\cos \alpha +\cos \beta )}^{2}}+{{(\sin \alpha +\sin \beta )}^{2}}\]\[={{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +2\cos \alpha \cos \beta +{{\sin }^{2}}\alpha +\] \[{{\sin }^{2}}\beta +2\sin \alpha \sin \beta \] \[=2\{1+\cos (\alpha -\beta )\}=4{{\cos }^{2}}\left( \frac{\alpha -\beta }{2} \right)\].
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