If x + 1x - 1 = a/b and 1 - y1 + y = b/a , then the value of x - y1 + xy is?

Answer: Option D

\begin{aligned} & {\text{Given ,}} \\ & \frac{{x + 1}}{{x - 1}} = \frac{a}{b} \\ & \left( {{\text{Using componendo & dividendo}}} \right) \\ & \Leftrightarrow \frac{x}{1} = \frac{{a + b}}{{a - b}} \\ & \Leftrightarrow x = \frac{{a + b}}{{a - b}}\,.....(i) \\ & {\text{Again,}}\frac{{1 - y}}{{1 + y}} = \frac{b}{a} \\ & \Leftrightarrow \frac{{1 + y}}{{1 - y}} = \frac{a}{b} \\ & \Leftrightarrow \frac{1}{y} = \frac{{a + b}}{{a - b}} \\ & \Leftrightarrow y = \frac{{a - b}}{{a + b}}\,.....(ii) \\ & {\text{From question,}} \\ & \frac{{x - y}}{{1 + xy}} \\ & \Rightarrow \frac{{\frac{{a + b}}{{a - b}} - \frac{{a - b}}{{a + b}}}}{{1 + \left( {\frac{{a + b}}{{a - b}}} \right)\left( {\frac{{a - b}}{{a + b}}} \right)}} \\ & \Rightarrow \frac{{{{\left( {a + b} \right)}^2} - {{\left( {a - b} \right)}^2}}}{{\left( {{a^2} - {b^2}} \right)\left( {1 + 1} \right)}} \\ & \Rightarrow \frac{{4ab}}{{2\left( {{a^2} - {b^2}} \right)}} \\ & \Rightarrow \frac{{2ab}}{{{a^2} - {b^2}}} \\\end{aligned}