If `f(x)=log {(1+x)/(1-x)}`, then `f(x)+f(y)` is

Options

A. `f(x+y)`
B. `f{(x+y)/(1+xy)}`
C. `(x+y) f{(1)/(1+xy)}`
D. `(f(x)+f(y))/(1+xy)`

Answer

To solve the problem, we need to find the expression for \( f(x) + f(y) \) given that \( f(x) = \log \left( \frac{1+x}{1-x} \right) \).

### Step-by-Step Solution:

1. **Write down the functions**:
\[ f(x) = \log \left( \frac{1+x}{1-x} \right) \]
\[ f(y) = \log \left( \frac{1+y}{1-y} \right) \]

2. **Add the two functions**:
\[ f(x) + f(y) = \log \left( \frac{1+x}{1-x} \right) + \log \left( \frac{1+y}{1-y} \right) \]

3. **Use the property of logarithms**:
The property of logarithms states that \( \log(a) + \log(b) = \log(ab) \). Therefore, we can combine the two logarithms:
\[ f(x) + f(y) = \log \left( \frac{1+x}{1-x} \cdot \frac{1+y}{1-y} \right) \]

4. **Simplify the product inside the logarithm**:
\[ = \log \left( \frac{(1+x)(1+y)}{(1-x)(1-y)} \right) \]

5. **Expand the numerator and denominator**:
- The numerator:
\[ (1+x)(1+y) = 1 + x + y + xy \]
- The denominator:
\[ (1-x)(1-y) = 1 - x - y + xy \]

6. **Combine the results**:
Thus, we have:
\[ f(x) + f(y) = \log \left( \frac{1 + x + y + xy}{1 - x - y + xy} \right) \]

7. **Recognize the form of \( f \)**:
Notice that the expression inside the logarithm resembles \( f \left( \frac{x+y}{1+xy} \right) \):
\[ f\left( \frac{x+y}{1+xy} \right) = \log \left( \frac{1 + \frac{x+y}{1+xy}}{1 - \frac{x+y}{1+xy}} \right) \]

8. **Simplify \( f\left( \frac{x+y}{1+xy} \right) \)**:
- The numerator becomes:
\[ 1 + \frac{x+y}{1+xy} = \frac{1 + xy + x + y}{1 + xy} \]
- The denominator becomes:
\[ 1 - \frac{x+y}{1+xy} = \frac{1 + xy - x - y}{1 + xy} \]

9. **Final expression**:
Therefore, we can conclude that:
\[ f(x) + f(y) = f\left( \frac{x+y}{1+xy} \right) \]

### Conclusion:
Thus, \( f(x) + f(y) = f\left( \frac{x+y}{1+xy} \right) \).