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? Options: A . &nbsp$$\frac{{{a^2} - {b^2}}}{{ab}}$$ B . &nbsp$$\frac{{{a^2} + {b^2}}}{{2ab}}$$ C . &nbsp$$\frac{{{a^2} - {b^2}}}{{2ab}}$$ D . &nbsp$$\frac{{2ab}}{{{a^2} - {b^2}}}$$

If $$\frac{{x + 1}}{{x - 1}} = \frac{a}{b}$$ and $$\frac{{1

Question

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?

Options:

A . $$\frac{{{a^2} - {b^2}}}{{ab}}$$

B . $$\frac{{{a^2} + {b^2}}}{{2ab}}$$

C . $$\frac{{{a^2} - {b^2}}}{{2ab}}$$

D . $$\frac{{2ab}}{{{a^2} - {b^2}}}$$

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

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