If in a `!ABC ,a tan A + btanB =(a + b) tan((A+B)/2)` , then

Options

A. A = B
B. A = -B
C. A = 2B
D. B = 2A

Answer

To solve the problem, we need to prove that if in triangle \(ABC\), the equation \(a \tan A + b \tan B = (a + b) \tan\left(\frac{A + B}{2}\right)\) holds, then it implies that angles \(A\) and \(B\) are equal, which means triangle \(ABC\) is isosceles.

### Step-by-Step Solution:

1. **Start with the given equation**:
\[ a \tan A + b \tan B = (a + b) \tan\left(\frac{A + B}{2}\right) \]

2. **Rewrite the left-hand side**:
\[ a \tan A + b \tan B = a \frac{\sin A}{\cos A} + b \frac{\sin B}{\cos B} \]

3. **Express \(\tan\left(\frac{A + B}{2}\right)\)** using the tangent addition formula:
\[ \tan\left(\frac{A + B}{2}\right) = \frac{\sin\left(\frac{A + B}{2}\right)}{\cos\left(\frac{A + B}{2}\right)} \]

4. **Substituting back into the equation**:
\[ a \frac{\sin A}{\cos A} + b \frac{\sin B}{\cos B} = (a + b) \frac{\sin\left(\frac{A + B}{2}\right)}{\cos\left(\frac{A + B}{2}\right)} \]

5. **Cross-multiply to eliminate the fractions**:
\[ a \sin A \cos\left(\frac{A + B}{2}\right) + b \sin B \cos\left(\frac{A + B}{2}\right) = (a + b) \sin\left(\frac{A + B}{2}\right) \cos A \cos B \]

6. **Rearranging the equation**:
\[ a \sin A \cos\left(\frac{A + B}{2}\right) - (a + b) \sin\left(\frac{A + B}{2}\right) \cos A \cos B + b \sin B \cos\left(\frac{A + B}{2}\right) = 0 \]

7. **Using the sine subtraction formula**:
\[ \sin A - \sin\left(\frac{A + B}{2}\right) = \sin B - \sin\left(\frac{A + B}{2}\right) \]

8. **From the above, conclude that**:
\[ \sin A = \sin B \]

9. **Since \(A\) and \(B\) are angles in a triangle, we conclude that**:
\[ A = B \]

10. **Thus, triangle \(ABC\) is isosceles**.

### Final Conclusion:
If \(a \tan A + b \tan B = (a + b) \tan\left(\frac{A + B}{2}\right)\), then angles \(A\) and \(B\) are equal, which means triangle \(ABC\) is isosceles.