A) \[-2\sin (\alpha +\beta )\]
B) \[-2\cos (\alpha +\beta )\]
C) \[2\sin (\alpha +\beta )\]
D) \[2\cos (\alpha +\beta )\]
Correct Answer: B
Solution :
[b] \[{{(cos\alpha +cos\beta )}^{2}}-{{(sin\alpha +sin\beta )}^{2}}=0\] Or \[(co{{s}^{2}}\alpha +co{{s}^{2}}\beta +2\cos \alpha \cos \beta )\] \[-(si{{n}^{2}}\alpha +si{{n}^{2}}\beta +2\sin \alpha \sin \beta )=0\] Or \[\cos 2\alpha +\cos 2\beta =-2(cos\alpha \cos \beta -\sin \alpha \sin \beta )\] \[=-2\cos (\alpha +\beta )\]
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