If tan A =1 and tan b = ` sqrt3` , evaluate :
sin A cos B+ cos A sin B.

Answer

To solve the problem, we need to evaluate the expression \( \sin A \cos B + \cos A \sin B \) given that \( \tan A = 1 \) and \( \tan B = \sqrt{3} \).

### Step-by-Step Solution:

1. **Determine the angles A and B**:
- Since \( \tan A = 1 \), we know that \( A = 45^\circ \) (because \( \tan 45^\circ = 1 \)).
- Since \( \tan B = \sqrt{3} \), we know that \( B = 60^\circ \) (because \( \tan 60^\circ = \sqrt{3} \)).

2. **Find the sine and cosine values**:
- For \( A = 45^\circ \):
- \( \sin A = \sin 45^\circ = \frac{1}{\sqrt{2}} \)
- \( \cos A = \cos 45^\circ = \frac{1}{\sqrt{2}} \)
- For \( B = 60^\circ \):
- \( \sin B = \sin 60^\circ = \frac{\sqrt{3}}{2} \)
- \( \cos B = \cos 60^\circ = \frac{1}{2} \)

3. **Substitute the values into the expression**:
- We need to evaluate \( \sin A \cos B + \cos A \sin B \):
\[ \sin A \cos B + \cos A \sin B = \left( \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \right) + \left( \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} \right) \]

4. **Calculate each term**:
- First term:
\[ \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{1}{2\sqrt{2}} \]
- Second term:
\[ \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2\sqrt{2}} \]

5. **Combine the terms**:
- Now, we combine the two terms:
\[ \frac{1}{2\sqrt{2}} + \frac{\sqrt{3}}{2\sqrt{2}} = \frac{1 + \sqrt{3}}{2\sqrt{2}} \]

6. **Simplify the expression**:
- To simplify, we can multiply the numerator and the denominator by \( \sqrt{2} \):
\[ \frac{(1 + \sqrt{3})\sqrt{2}}{2 \cdot 2} = \frac{(1 + \sqrt{3})\sqrt{2}}{4} \]

### Final Answer:
\[ \frac{(1 + \sqrt{3})\sqrt{2}}{4} \]