Question

If $ xy+yz+zx=1, $ then $ \sum{\frac{x+y}{1-xy}} $ is equal to:

• A. $ \frac{4}{xyz} $
• B. $ \frac{1}{xyz} $
• C. $ ~xyz $
• D. none of these

Answer 1

Answer: (B)

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} $