If (1 + tanA)(1 + tan B) = 2, then find the value of (A +B).

Answer

To solve the equation \((1 + \tan A)(1 + \tan B) = 2\) and find the value of \(A + B\), we can follow these steps:

### Step 1: Expand the Left-Hand Side
We start with the equation:
\[ (1 + \tan A)(1 + \tan B) = 2 \]
Expanding the left-hand side, we get:
\[ 1 + \tan A + \tan B + \tan A \tan B = 2 \]

### Step 2: Rearrange the Equation
Now, we can rearrange the equation to isolate the terms involving \(\tan A\) and \(\tan B\):
\[ \tan A + \tan B + \tan A \tan B = 2 - 1 \]
This simplifies to:
\[ \tan A + \tan B + \tan A \tan B = 1 \]

### Step 3: Use the Tangent Addition Formula
We know from the tangent addition formula:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
From our equation, we can substitute \(\tan A + \tan B\) and \(\tan A \tan B\):
Let \(x = \tan A\) and \(y = \tan B\). Then:
\[ x + y + xy = 1 \]
This can be rearranged to:
\[ x + y = 1 - xy \]

### Step 4: Substitute into the Tangent Addition Formula
Now substituting into the tangent addition formula:
\[ \tan(A + B) = \frac{x + y}{1 - xy} \]
Substituting \(x + y = 1 - xy\):
\[ \tan(A + B) = \frac{1 - xy}{1 - xy} \]
This simplifies to:
\[ \tan(A + B) = 1 \]

### Step 5: Solve for \(A + B\)
The equation \(\tan(A + B) = 1\) implies:
\[ A + B = \frac{\pi}{4} + n\pi \quad \text{for } n \in \mathbb{Z} \]
For the principal value, we take \(n = 0\):
\[ A + B = \frac{\pi}{4} \]

### Final Answer
Thus, the value of \(A + B\) is:
\[ \boxed{\frac{\pi}{4}} \]