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)]
$=\frac{1}{xyz}$