A) 0
B) - 1
C) 1
D) None of these
Correct Answer: C
Solution :
[c] : \[{{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +{{\sin }^{2}}\gamma \] \[=\frac{{{\tan }^{2}}\alpha }{1+{{\tan }^{2}}\alpha }+\frac{{{\tan }^{2}}\beta }{1+{{\tan }^{2}}\beta }+\frac{{{\tan }^{2}}\gamma }{1+{{\tan }^{2}}\gamma }\] \[=\frac{x}{1+x}+\frac{y}{1+y}+\frac{z}{1+z}\] (Here,\[x={{\tan }^{2}}\alpha ,y={{\tan }^{2}}\beta ,z={{\tan }^{2}}\gamma \]) \[=\frac{(x+y+z)+(xy+yz+zx+2xyz)+xy+yz+zx+xyz}{(1+x)(1+y)(1+z)}\]\[(\because xy+yz+zx+2xyz=1)\] \[=\frac{1+x+y+z+xy+yz+zx+xyz}{(1+x)(1+y)(1+z)}=1\]
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