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

Answer

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

### Step-by-Step Solution:

1. **Identify 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 Cosine and Sine Values:**
- For \( A = 45^\circ \):
- \( \cos A = \cos 45^\circ = \frac{1}{\sqrt{2}} \)
- \( \sin A = \sin 45^\circ = \frac{1}{\sqrt{2}} \)
- For \( B = 60^\circ \):
- \( \cos B = \cos 60^\circ = \frac{1}{2} \)
- \( \sin B = \sin 60^\circ = \frac{\sqrt{3}}{2} \)

3. **Substitute Values into the Expression:**
- Now substitute the values into the expression \( \cos A \cos B - \sin A \sin B \):
\[ \cos A \cos B - \sin A \sin B = \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{2}\right) - \left(\frac{1}{\sqrt{2}}\right) \left(\frac{\sqrt{3}}{2}\right) \]

4. **Calculate Each Term:**
- The first term:
\[ \cos A \cos B = \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{1}{2\sqrt{2}} \]
- The second term:
\[ \sin A \sin B = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2\sqrt{2}} \]

5. **Combine the Terms:**
- Now, combine the two terms:
\[ \cos A \cos B - \sin A \sin B = \frac{1}{2\sqrt{2}} - \frac{\sqrt{3}}{2\sqrt{2}} = \frac{1 - \sqrt{3}}{2\sqrt{2}} \]

### Final Answer:
\[ \cos A \cos B - \sin A \sin B = \frac{1 - \sqrt{3}}{2\sqrt{2}} \]