If ` xy + yz + zx = 1 ` then the value of ` (1 + y ^(2))/(( x + y) ( y + z))` is

Options

A. 2
B. 3
C. 4
D. 1

Answer

To solve the problem, we need to find the value of the expression \( \frac{1 + y^2}{(x + y)(y + z)} \) given that \( xy + yz + zx = 1 \).

### Step-by-Step Solution:

1. **Given Equation**: We start with the equation provided:
\[ xy + yz + zx = 1 \]

2. **Assumption**: Since we have three variables and only one equation, we can assume a value for one of the variables to simplify our calculations. Let's set \( z = 0 \).

3. **Substituting z = 0**: Substituting \( z = 0 \) into the equation gives:
\[ xy + y(0) + (0)x = 1 \implies xy = 1 \]

4. **Choosing Values for x and y**: From \( xy = 1 \), we can choose \( x = 1 \) and \( y = 1 \) (since \( 1 \times 1 = 1 \)).

5. **Substituting Values into the Expression**: Now we substitute \( x = 1 \), \( y = 1 \), and \( z = 0 \) into the expression we want to evaluate:
\[ \frac{1 + y^2}{(x + y)(y + z)} = \frac{1 + 1^2}{(1 + 1)(1 + 0)} \]

6. **Calculating the Numerator**: The numerator becomes:
\[ 1 + 1^2 = 1 + 1 = 2 \]

7. **Calculating the Denominator**: The denominator becomes:
\[ (1 + 1)(1 + 0) = 2 \times 1 = 2 \]

8. **Final Calculation**: Now we can compute the value of the entire expression:
\[ \frac{2}{2} = 1 \]

### Conclusion:
Thus, the value of the expression \( \frac{1 + y^2}{(x + y)(y + z)} \) is:
\[ \boxed{1} \]