If `sin A=sqrt3/2 and cos B = sqrt3/2` , find the value of : `(tanA-tanB)/(1+tanAtanB)`

Answer

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

### Step-by-Step Solution:

1. **Identify Angles A and B**:
- Given \(\sin A = \frac{\sqrt{3}}{2}\), we know that this corresponds to \(A = 60^\circ\) (since \(\sin 60^\circ = \frac{\sqrt{3}}{2}\)).
- Given \(\cos B = \frac{\sqrt{3}}{2}\), we know that this corresponds to \(B = 30^\circ\) (since \(\cos 30^\circ = \frac{\sqrt{3}}{2}\)).

2. **Calculate \(\tan A\) and \(\tan B\)**:
- For \(A = 60^\circ\), \(\tan A = \tan 60^\circ = \sqrt{3}\).
- For \(B = 30^\circ\), \(\tan B = \tan 30^\circ = \frac{1}{\sqrt{3}}\).

3. **Substitute Values into the Expression**:
- Now substitute the values of \(\tan A\) and \(\tan B\) into the expression:
\[ \frac{\tan A - \tan B}{1 + \tan A \tan B} = \frac{\sqrt{3} - \frac{1}{\sqrt{3}}}{1 + \sqrt{3} \cdot \frac{1}{\sqrt{3}}} \]

4. **Simplify the Denominator**:
- The denominator simplifies as follows:
\[ 1 + \sqrt{3} \cdot \frac{1}{\sqrt{3}} = 1 + 1 = 2 \]

5. **Simplify the Numerator**:
- The numerator simplifies as follows:
\[ \sqrt{3} - \frac{1}{\sqrt{3}} = \frac{3}{\sqrt{3}} - \frac{1}{\sqrt{3}} = \frac{3 - 1}{\sqrt{3}} = \frac{2}{\sqrt{3}} \]

6. **Combine the Results**:
- Now substitute back into the expression:
\[ \frac{\frac{2}{\sqrt{3}}}{2} = \frac{2}{\sqrt{3} \cdot 2} = \frac{1}{\sqrt{3}} \]

7. **Rationalize the Denominator**:
- To rationalize the denominator, multiply the numerator and denominator by \(\sqrt{3}\):
\[ \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3} \]

### Final Answer:
Thus, the final value is \(\frac{\sqrt{3}}{3}\).