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

Solution(By Examveda Team)

$$\eqalign{ & \frac{{x + 1}}{{x - 1}}{\text{ = }}\frac{a}{b}{\text{ }} \cr & {\text{By componendo and dividendo}} \cr & \frac{{x + 1 + x - 1}}{{x + 1 - x + 1}} = \frac{{a + b}}{{a - b}} \cr & \Rightarrow \frac{{2x}}{2} = \frac{{a + b}}{{a - b}} \cr & \Rightarrow x = \frac{{a + b}}{{a - b}}.....(i) \cr & {\text{Again ,}}\frac{{1 - y}}{{1 + y}}{\text{ = }}\frac{b}{a}{\text{ }} \cr & \Rightarrow \frac{{1 + y}}{{1 - y}} = \frac{a}{b} \cr & \Rightarrow \frac{{1 + y + 1 - y}}{{1 + y - 1 + y}} = \frac{{a + b}}{{a - b}} \cr & \Rightarrow \frac{2}{{2y}} = \frac{{a + b}}{{a - b}} \cr & \Rightarrow \frac{1}{y} = \frac{{a + b}}{{a - b}} \cr & \Rightarrow y = \frac{{a - b}}{{a + b}}.....(ii) \cr & {\text{Subtracting equation (ii) from (i) we get }} \cr & \therefore x - y = \frac{{a + b}}{{a - b}} - \frac{{a - b}}{{a + b}} \cr & \left[ {\because {{\left( {a + b} \right)}^2} - {{\left( {a - b} \right)}^2} = 4ab\,\,\,and\,\,{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)} \right] \cr & = \frac{{{{\left( {a + b} \right)}^2} - {{\left( {a - b} \right)}^2}}}{{\left( {a - b} \right)\left( {a + b} \right)}} \cr & = \frac{{4ab}}{{{a^2} - {b^2}}} \cr & {\text{Multiply equation (i) and (ii) we get}} \cr & xy = \frac{{a + b}}{{a - b}} \times \frac{{a - b}}{{a + b}} = 1 \cr & \therefore {\text{Expression,}} \cr & {\text{ = }}\frac{{x - y}}{{1 + xy}} = \frac{{\frac{{4ab}}{{{a^2} - {b^2}}}}}{{1 + 1}} \cr &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{4ab}}{{2\left( {{a^2} - {b^2}} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2ab}}{{{a^2} - {b^2}}} \cr} $$