A) \[2\cos \alpha \cos \beta \cos \gamma \]
B) \[-\cos \alpha \cos \beta \cos \gamma \]
C) \[+\cos \alpha \cos \beta \cos \gamma \]
D) \[+2sin\alpha sin\beta sin\gamma \]
Correct Answer: C
Solution :
If \[(1+\sin \alpha )(1+\sin \beta )(1+sin\gamma )=k\] And \[(1-\sin \alpha )\,(1-\sin \beta )\,(1-\sin \gamma )=k\] The the value of \[{{k}^{2}}=k.k.\] \[=(1+\sin \alpha )(1+sin\beta )(1+sin\gamma )(1-sin\alpha )\]\[(1-\sin \beta )(1-sin\gamma )\] \[=(1-{{\sin }^{2}}\alpha )(1-{{\sin }^{2}}\beta )(1-{{\sin }^{2}}\gamma )\] \[\Rightarrow \,\,{{k}^{2}}={{\cos }^{2}}\alpha {{\cos }^{2}}\beta {{\cos }^{2}}\gamma \] \[\therefore \,\,\,k=+\cos \alpha cos\beta cos\gamma .\]
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