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. **Understand the Given Condition**:
We start with the condition \(xy + yz + zx = 1\). This is a key relationship that we will use to simplify our expression.

2. **Rewrite the Expression**:
We need to evaluate the expression \(\frac{1 + y^2}{(x+y)(y+z)}\). Let's first expand the denominator:
\[ (x+y)(y+z) = xy + xz + y^2 + yz \]

3. **Substitute the Condition**:
From the condition \(xy + yz + zx = 1\), we can express \(xy + yz\) in terms of \(1 - zx\):
\[ xy + yz = 1 - zx \]
Therefore, we can substitute this into our expression for the denominator:
\[ (x+y)(y+z) = (1 - zx) + xz + y^2 = 1 + y^2 - zx \]

4. **Rewrite the Expression Again**:
Now, substituting back into our expression, we have:
\[ \frac{1 + y^2}{(x+y)(y+z)} = \frac{1 + y^2}{1 + y^2 - zx} \]

5. **Simplify the Expression**:
Since \(xy + yz + zx = 1\), we can see that \(zx\) can be expressed as \(1 - (xy + yz)\). Thus, we can rewrite the denominator as:
\[ 1 + y^2 - zx = 1 + y^2 - (1 - (xy + yz)) = y^2 + xy + yz \]
However, we notice that this is simply \(1\) because of our original condition. Thus:
\[ \frac{1 + y^2}{1} = 1 + y^2 \]

6. **Final Value**:
Since we have simplified our expression to \(1\), we conclude that:
\[ \frac{1 + y^2}{(x+y)(y+z)} = 1 \]

### Conclusion:
The value of the expression \(\frac{1 + y^2}{(x+y)(y+z)}\) is \(1\).