If `(x + 1)/( x - 1) = (a)/(b) and (1 - y)/( 1 + y) = (b)/(a)` then the value of `( x - y)/( 1 + xy)` is

Options

A. `(2 ab)/( a^(2) - b^(2))`
B. ` (a^(2) - b^(2))/(2ab)`
C. `(a^(2) + b^(2))/( 2ab)`
D. `(a^(2) - b^(2))/( ab)`

Answer

To solve the given problem, we need to manipulate the equations step by step.

We are given two equations:

1. \(\frac{x + 1}{x - 1} = \frac{a}{b}\)
2. \(\frac{1 - y}{1 + y} = \frac{b}{a}\)

We want to find the value of \(\frac{x - y}{1 + xy}\).

### Step 1: Express \(x\) in terms of \(a\) and \(b\)

From the first equation:

\[ \frac{x + 1}{x - 1} = \frac{a}{b} \]

Cross-multiplying gives:

\[ b(x + 1) = a(x - 1) \]

Expanding both sides:

\[ bx + b = ax - a \]

Rearranging terms:

\[ bx - ax = -a - b \]

Factoring out \(x\):

\[ (b - a)x = -a - b \]

Thus, we can express \(x\) as:

\[ x = \frac{-a - b}{b - a} \]

### Step 2: Express \(y\) in terms of \(a\) and \(b\)

From the second equation:

\[ \frac{1 - y}{1 + y} = \frac{b}{a} \]

Cross-multiplying gives:

\[ a(1 - y) = b(1 + y) \]

Expanding both sides:

\[ a - ay = b + by \]

Rearranging terms:

\[ a - b = ay + by \]

Factoring out \(y\):

\[ a - b = y(a + b) \]

Thus, we can express \(y\) as:

\[ y = \frac{a - b}{a + b} \]

### Step 3: Substitute \(x\) and \(y\) into \(\frac{x - y}{1 + xy}\)

Now we need to calculate \(x - y\) and \(1 + xy\).

1. **Calculate \(x - y\)**:

\[ x - y = \frac{-a - b}{b - a} - \frac{a - b}{a + b} \]

Finding a common denominator:

\[ x - y = \frac{(-a - b)(a + b) - (a - b)(b - a)}{(b - a)(a + b)} \]

Expanding the numerator:

\[ = \frac{-a^2 - ab - ab - b^2 - (ab - a^2 + b^2 - ab)}{(b - a)(a + b)} \]

Simplifying:

\[ = \frac{-a^2 - 2ab - b^2 + a^2 + b^2}{(b - a)(a + b)} = \frac{-2ab}{(b - a)(a + b)} \]

2. **Calculate \(xy\)**:

\[ xy = \left(\frac{-a - b}{b - a}\right) \left(\frac{a - b}{a + b}\right) \]

Simplifying:

\[ = \frac{(-a - b)(a - b)}{(b - a)(a + b)} = \frac{-(a^2 - ab + ab - b^2)}{(b - a)(a + b)} = \frac{-(a^2 - b^2)}{(b - a)(a + b)} \]

Thus,

\[ 1 + xy = 1 - \frac{(a^2 - b^2)}{(b - a)(a + b)} = \frac{(b - a)(a + b) - (a^2 - b^2)}{(b - a)(a + b)} \]

The numerator simplifies to:

\[ = \frac{(b - a)(a + b) - (a^2 - b^2)}{(b - a)(a + b)} = \frac{(b - a)(a + b) - (a - b)(a + b)}{(b - a)(a + b)} = \frac{(b - a)(a + b) + (b - a)(a + b)}{(b - a)(a + b)} = \frac{2(b - a)(a + b)}{(b - a)(a + b)} = 2 \]

### Step 4: Combine results to find \(\frac{x - y}{1 + xy}\)

Now we can find:

\[ \frac{x - y}{1 + xy} = \frac{\frac{-2ab}{(b - a)(a + b)}}{2} = \frac{-ab}{(b - a)(a + b)} \]

### Final Answer

Thus, the value of \(\frac{x - y}{1 + xy}\) is:

\[ \frac{-ab}{(b - a)(a + b)} \]