A) 2
B) -1
C) -2
D) 0
Correct Answer: A
Solution :
We have, \[{{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +{{\sin }^{2}}\gamma -2\cos \alpha \cos \beta \cos \gamma \] = \[3-[{{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma ]-2\cos \alpha \cos \beta \cos \gamma \] = \[3-\left[ \frac{1+\cos 2\alpha }{2}+\frac{1+\cos 2\beta }{2}+\frac{1+\cos 2\gamma }{2} \right]\] \[-2\cos \alpha \cos \beta \cos \gamma \] \[=3-\frac{1}{2}[3+\cos 2\alpha +\cos 2\beta +\cos 2\gamma ]-2\cos \alpha \cos \beta \cos \gamma \] \[=3-\frac{3}{2}-\frac{1}{2}(\cos 2\alpha +\cos 2\beta )-\frac{1}{2}\cos 2\gamma -2\cos \alpha \cos \beta \cos \gamma \] = \[\frac{3}{2}-\frac{1}{2}[-2\cos \gamma \cos (\alpha -\beta )]-\frac{1}{2}[2{{\cos }^{2}}\gamma -1]\] \[-2\cos \alpha \cos \beta \cos \gamma \] =\[\frac{3}{2}+\cos \gamma \cos (\alpha -\beta )-{{\cos }^{2}}\gamma +\frac{1}{2}-2\cos \alpha \cos \beta \cos \gamma \] = 2.
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