If `x=(p+q)/(p-q)` and `y=(p-q)/(p+q)` then the value of `(x-y)/(x+y)` is

Options

A. `(p^(2)+q^(2))/(2pq)`
B. `(2pq)/(p^(2)+q^(2))`
C. `(2pq)/(p^(2)-q^(2))`
D. `(2(p^(2)-q^(2)))/(pq)`

Answer

To solve the problem, we need to find the value of \((x - y) / (x + y)\) given that \(x = \frac{p + q}{p - q}\) and \(y = \frac{p - q}{p + q}\).

### Step-by-Step Solution:

1. **Find \(x - y\)**:
\[ x - y = \frac{p + q}{p - q} - \frac{p - q}{p + q} \]
To subtract these fractions, we need a common denominator, which is \((p - q)(p + q)\):
\[ x - y = \frac{(p + q)^2 - (p - q)^2}{(p - q)(p + q)} \]

2. **Expand the numerators**:
\[ (p + q)^2 = p^2 + 2pq + q^2 \]
\[ (p - q)^2 = p^2 - 2pq + q^2 \]
Now substituting these back:
\[ x - y = \frac{(p^2 + 2pq + q^2) - (p^2 - 2pq + q^2)}{(p - q)(p + q)} \]
Simplifying the numerator:
\[ x - y = \frac{4pq}{(p - q)(p + q)} \]

3. **Find \(x + y\)**:
\[ x + y = \frac{p + q}{p - q} + \frac{p - q}{p + q} \]
Again, we need a common denominator:
\[ x + y = \frac{(p + q)^2 + (p - q)^2}{(p - q)(p + q)} \]

4. **Expand the numerators**:
Using the expansions from earlier:
\[ x + y = \frac{(p^2 + 2pq + q^2) + (p^2 - 2pq + q^2)}{(p - q)(p + q)} \]
Simplifying the numerator:
\[ x + y = \frac{2p^2 + 2q^2}{(p - q)(p + q)} = \frac{2(p^2 + q^2)}{(p - q)(p + q)} \]

5. **Now, find \(\frac{x - y}{x + y}\)**:
\[ \frac{x - y}{x + y} = \frac{\frac{4pq}{(p - q)(p + q)}}{\frac{2(p^2 + q^2)}{(p - q)(p + q)}} \]
The \((p - q)(p + q)\) cancels out:
\[ \frac{x - y}{x + y} = \frac{4pq}{2(p^2 + q^2)} = \frac{2pq}{p^2 + q^2} \]

### Final Answer:
\[ \frac{x - y}{x + y} = \frac{2pq}{p^2 + q^2} \]