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

Given : $$\frac{x+1}{x-1}=\frac{a}{b}$$

=> $$b(x+1)=a(x-1)$$

=> $$bx+b=ax-a$$

=> $$x(a-b)=a+b$$

=> $$x=\frac{a+b}{a-b}$$

Similarly, $$y=\frac{a-b}{a+b}$$

To find : $$\frac{x-y}{1+xy}$$

= $$[(\frac{a+b}{a-b})-(\frac{a-b}{a+b})]\div[1+(\frac{a+b}{a-b})(\frac{a-b}{a+b})]$$

= $$(\frac{(a+b)^2-(a-b)^2}{(a+b)(a-b)})\div(1+1)$$

= $$(\frac{(a^2+b^2+2ab)-(a^2+b^2-2ab)}{a^2-b^2})\times(\frac{1}{2})$$

= $$\frac{4ab}{2(a^2-b^2)}$$

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

=> Ans - (C)