A) \[-\,\,1\]
B) 2
C) 1
D) 0
Correct Answer: D
Solution :
\[{{\sin }^{2}}x+{{\sin }^{2}}y+{{\sin }^{2}}z={{(\sin x+\sin y+\sin z)}^{2}}\] |
\[[\because {{(a+b+c)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\,\,(ab+bc+ca)]\] |
\[\Rightarrow \]\[\sin x\sin y+\sin y\sin \,\,z+sin\,\,z\,\,sin\,\,x=0\] |
Dividing both sides by \[x\sin y\sin z,\] |
\[\Rightarrow \] \[\frac{1}{\sin x}+\frac{1}{\sin y}+\frac{1}{\sin \,\,z}=0\] |
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