If $$\frac{{x + 1}}{{x - 1}} = \frac{a}{b}$$   and $$\frac{{1 - y}}{{1 + y}} = \frac{b}{a}{\text{,}}$$   then the value of $$\frac{{x - y}}{{1 + xy}}$$   is?

Explanation:-

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