Question : 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:
Option 1: $\frac{2ab}{a^{2}-b^{2}}$
Option 2: $\frac{a^{2}-b^{2}}{2ab}$
Option 3: $\frac{a^{2}+b^{2}}{2ab}$
Option 4: $\frac{a^{2}-b^{2}}{ab}$
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Correct Answer: $\frac{2ab}{a^{2}-b^{2}}$
Solution :
$\frac{x+1}{x-1}=\frac{a}{b}$
⇒ $bx+b = ax-a$
⇒ $x = \frac{a+b}{a-b}$
Multiplying and dividing by $(a+b)$ in RHS, we get:
⇒ $x=\frac{(a+b)^2}{a^2-b^2}$ ----------------(i)
Again, $\frac{1-y}{1+y}=\frac{b}{a}$
⇒ $a-ay = b+by$
⇒ $y=\frac{a-b}{a+b}$
Multiplying and diving by $(a-b)$ in RHS, we get:
⇒ $y=\frac{(a-b)^2}{a^2-b^2}$----------------(ii)
From (i) and (ii),
$\frac{x-y}{1+xy}$ = $\frac{\frac{(a+b)^2}{a^2-b^2}-\frac{(a-b)^2}{a^2-b^2}}{1+[\frac{(a+b)^2}{a^2-b^2}×\frac{(a-b)^2}{a^2-b^2}]}$
= $\frac{\frac{(a+b)^2-(a-b)^2}{a^2-b^2}}{1+\frac{(a^2-b^2)^2}{(a^2-b^2)^2}}$
= $\frac{\frac{4ab}{a^2-b^2}}{2}$
= $\frac{2ab}{a^{2}-b^{2}}$
Hence, the correct answer is $\frac{2ab}{a^{2}-b^{2}}$.
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