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

1 $\frac{a^{2} - b^{2}}{a b}$ $\frac{{{a^2} - {b^2}}}{{ab}}$
2 $\frac{a^{2} + b^{2}}{2 a b}$ $\frac{{{a^2} + {b^2}}}{{2ab}}$
3 $\frac{a^{2} - b^{2}}{2 a b}$ $\frac{{{a^2} - {b^2}}}{{2ab}}$
4 $\frac{2 a b}{a^{2} - b^{2}}$ $\frac{{2ab}}{{{a^2} - {b^2}}}$

Answer:- 1

Explanation:-

Solution:

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

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