A) \[\frac{4}{xyz}\]
B) \[\frac{1}{xyz}\]
C) \[~xyz\]
D) none of these
Correct Answer: B
Solution :
Since, \[xy+yz+zx=1\] Put, \[x=\cot A,\,y=\cot B,z=\cot C\] \[\Rightarrow \]\[\cot A\cot B+\cot C[\cot \,B+\cot A]=1\]?(i) \[\Rightarrow \] \[\cot C[\cot A+\cot B]=1-\cot A\cot B\] \[\Rightarrow \] \[\cot C=\frac{1-\cot A\cot B}{\cot A+\cot B}\] \[\Rightarrow \] \[\frac{\cot A+\cot B}{1-\cot A\cot B}=\frac{1}{\cot C}\] \[\therefore \] \[\sum{\frac{x+y}{1-xy}}=\sum{\frac{\cot A+\cot B}{1-\cot A\cot B}}\] \[=\sum{\frac{1}{\cot \,C}}\] \[=\frac{1}{\cot C}+\frac{1}{\cot A}+\frac{1}{\cot B}\] \[=\frac{\cot A\cot B+\cot \,B\,\cot \,C+\cot \,A\,\cot \,C}{\cot A\,\cot B\cot C}\] \[=\frac{1}{\cot A\cot B\cot C}\] [From (i)] \[=\frac{1}{xyz}\]
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