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

### Concept & Formula The core mathematical tool for this problem is the rule of **Componendo and Dividendo**, which states that if $\frac{p}{q} = \frac{r}{s}$, then $\frac{p+q}{p-q} = \frac{r+s}{r-s}$. This property allows us to instantly untangle variables trapped in $(+1)$ and $(-1)$ fractional relationships. ### Step-by-Step Solution **Given:** 1) $\frac{x+1}{x-1} = \frac{a}{b}$ 2) $\frac{1-y}{1+y} = \frac{b}{a}$ **Calculation:** Step 1: Apply Componendo & Dividendo to the first equation. $$ \frac{(x+1) + (x-1)}{(x+1) - (x-1)} = \frac{a+b}{a-b} $$ $$ \frac{2x}{2} = \frac{a+b}{a-b} \implies x = \frac{a+b}{a-b} $$ Step 2: Apply Componendo & Dividendo to the second equation. $$ \frac{(1-y) + (1+y)}{(1-y) - (1+y)} = \frac{b+a}{b-a} $$ $$ \frac{2}{-2y} = \frac{a+b}{-(a-b)} \implies -\frac{1}{y} = -\frac{a+b}{a-b} \implies y = \frac{a-b}{a+b} $$ Step 3: Analyze the relationship between $x$ and $y$. Notice that $x = \frac{a+b}{a-b}$ and $y = \frac{a-b}{a+b}$. They are perfect reciprocals! Therefore, $xy = 1$. Step 4: Substitute into the target expression $\frac{x-y}{1+xy}$. Since $xy = 1$, the denominator is $1 + 1 = 2$. So we just need to find $\frac{x-y}{2}$. Step 5: Calculate $(x - y)$. $$ x - y = \frac{a+b}{a-b} - \frac{a-b}{a+b} = \frac{(a+b)^2 - (a-b)^2}{(a-b)(a+b)} $$ Using the algebraic identity $(a+b)^2 - (a-b)^2 = 4ab$: $$ x - y = \frac{4ab}{a^2 - b^2} $$ Step 6: Final substitution. $$ \frac{x-y}{1+xy} = \frac{\frac{4ab}{a^2 - b^2}}{2} = \frac{2ab}{a^2 - b^2} $$ ### Exam Strategy & Shortcut **The Value Substitution Method:** Instead of heavy algebra, substitute simple numbers. Let $a=3$ and $b=1$. From $\frac{x+1}{x-1} = 3 \implies 3x - 3 = x + 1 \implies 2x = 4 \implies x=2$. From $\frac{1-y}{1+y} = \frac{1}{3} \implies 1 + y = 3 - 3y \implies 4y = 2 \implies y = 0.5$. Target expression: $\frac{2 - 0.5}{1 + (2 \times 0.5)} = \frac{1.5}{2} = \frac{3}{4}$. Now, plug $a=3, b=1$ into Option A: $\frac{2(3)(1)}{3^2 - 1^2} = \frac{6}{8} = \frac{3}{4}$. It matches perfectly, bypassing all the algebraic manipulation! ### Common Pitfall When applying Componendo & Dividendo to $\frac{1-y}{1+y}$, students frequently mess up the negative signs, ending up with $y = \frac{a+b}{a-b}$ instead of its reciprocal. Careless sign errors in the denominator $(1-y) - (1+y) = -2y$ ruin the entire derivation. ### Final Answer **Therefore, the correct answer is $\frac{2ab}{a^2 - b^2}$.**