If `tan(A+B)=sqrt(3)andsin(A-B)=(1)/(2)`,then find the value of `tan (2A-3B)`.

Answer

To solve the problem, we need to find the value of \( \tan(2A - 3B) \) given that \( \tan(A + B) = \sqrt{3} \) and \( \sin(A - B) = \frac{1}{2} \).

### Step-by-Step Solution:

1. **Identify the angles from the given tangent value**:
\[ \tan(A + B) = \sqrt{3} \]
We know that \( \tan(60^\circ) = \sqrt{3} \). Therefore, we can conclude:
\[ A + B = 60^\circ \quad \text{(1)} \]

2. **Identify the angles from the given sine value**:
\[ \sin(A - B) = \frac{1}{2} \]
We know that \( \sin(30^\circ) = \frac{1}{2} \). Therefore, we can conclude:
\[ A - B = 30^\circ \quad \text{(2)} \]

3. **Solve the system of equations**:
Now we have two equations:
\[ A + B = 60^\circ \quad \text{(1)} \]
\[ A - B = 30^\circ \quad \text{(2)} \]
We can add these two equations:
\[ (A + B) + (A - B) = 60^\circ + 30^\circ \]
This simplifies to:
\[ 2A = 90^\circ \]
Dividing both sides by 2 gives:
\[ A = 45^\circ \]

4. **Substitute \( A \) back to find \( B \)**:
Now substitute \( A = 45^\circ \) into equation (1):
\[ 45^\circ + B = 60^\circ \]
Solving for \( B \):
\[ B = 60^\circ - 45^\circ = 15^\circ \]

5. **Calculate \( \tan(2A - 3B) \)**:
Now we need to find \( \tan(2A - 3B) \):
\[ 2A - 3B = 2(45^\circ) - 3(15^\circ) \]
This simplifies to:
\[ 90^\circ - 45^\circ = 45^\circ \]
Therefore:
\[ \tan(2A - 3B) = \tan(45^\circ) = 1 \]

### Final Answer:
\[ \tan(2A - 3B) = 1 \]