If `f(x)=log_e((1-x)/(1+x));` prove that `f(a)+f(b)=f((a+b)/(1+a b))`

Explanation

To prove that \( f(a) + f(b) = f\left(\frac{a+b}{1+ab}\right) \) where \( f(x) = \log_e\left(\frac{1-x}{1+x}\right) \), we will follow these steps:

### Step 1: Write the left-hand side
We start with the left-hand side of the equation:
\[ f(a) + f(b) = \log_e\left(\frac{1-a}{1+a}\right) + \log_e\left(\frac{1-b}{1+b}\right) \]

### Step 2: Use the property of logarithms
Using the property of logarithms that states \( \log(m) + \log(n) = \log(m \cdot n) \), we combine the two logarithms:
\[ f(a) + f(b) = \log_e\left(\frac{1-a}{1+a} \cdot \frac{1-b}{1+b}\right) \]

### Step 3: Simplify the expression inside the logarithm
Now, we simplify the expression inside the logarithm:
\[ \frac{1-a}{1+a} \cdot \frac{1-b}{1+b} = \frac{(1-a)(1-b)}{(1+a)(1+b)} \]

Expanding both the numerator and the denominator:
- Numerator:
\[ (1-a)(1-b) = 1 - a - b + ab \]
- Denominator:
\[ (1+a)(1+b) = 1 + a + b + ab \]

Thus, we have:
\[ f(a) + f(b) = \log_e\left(\frac{1 - a - b + ab}{1 + a + b + ab}\right) \]

### Step 4: Write the right-hand side
Now, we compute the right-hand side:
\[ f\left(\frac{a+b}{1+ab}\right) = \log_e\left(\frac{1 - \frac{a+b}{1+ab}}{1 + \frac{a+b}{1+ab}}\right) \]

### Step 5: Simplify the right-hand side
We simplify the expression inside the logarithm:
\[ 1 - \frac{a+b}{1+ab} = \frac{(1+ab) - (a+b)}{1+ab} = \frac{1 + ab - a - b}{1 + ab} \]
\[ 1 + \frac{a+b}{1+ab} = \frac{(1+ab) + (a+b)}{1+ab} = \frac{1 + ab + a + b}{1 + ab} \]

Thus, the right-hand side becomes:
\[ f\left(\frac{a+b}{1+ab}\right) = \log_e\left(\frac{1 + ab - a - b}{1 + ab + a + b}\right) \]

### Step 6: Compare both sides
Now we have:
- Left-hand side:
\[ f(a) + f(b) = \log_e\left(\frac{1 - a - b + ab}{1 + a + b + ab}\right) \]
- Right-hand side:
\[ f\left(\frac{a+b}{1+ab}\right) = \log_e\left(\frac{1 + ab - a - b}{1 + ab + a + b}\right) \]

Since both expressions inside the logarithms are equal, we conclude that:
\[ f(a) + f(b) = f\left(\frac{a+b}{1+ab}\right) \]

### Conclusion
Thus, we have proved that:
\[ f(a) + f(b) = f\left(\frac{a+b}{1+ab}\right) \]