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 $$\frac{{x - y}}{{1 + xy}}$$= ?

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

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

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

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

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

1.	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 $$\frac{{x - y}}{{1 + xy}}$$= ?
A.	$\frac{{2ab}}{{{a^2} - {b^2}}}$$
B.	$\frac{{{a^2} - {b^2}}}{{2ab}}$$
C.	$\frac{{{a^2} + {b^2}}}{{2ab}}$$
D.	$\frac{{{a^2} - {b^2}}}{{ab}}$$
Answer» B. $\frac{{{a^2} - {b^2}}}{{2ab}}$$