If tan (A + B) = `sqrt(3)` and tan (A - B) = 0, `0^(@) lt ` A + B `le 90^(@)` , find sin (A + B) and cos (A - B).

Answer

To solve the problem, we need to find the values of \( \sin(A + B) \) and \( \cos(A - B) \) given that \( \tan(A + B) = \sqrt{3} \) and \( \tan(A - B) = 0 \).

### Step-by-Step Solution:

1. **Identify the values of \( A + B \) and \( A - B \)**:
- We know that \( \tan(A + B) = \sqrt{3} \). The angle whose tangent is \( \sqrt{3} \) is \( 60^\circ \). Therefore, we can write:
\[ A + B = 60^\circ \]
- We also know that \( \tan(A - B) = 0 \). The angle whose tangent is \( 0 \) is \( 0^\circ \). Therefore, we can write:
\[ A - B = 0^\circ \]

2. **Set up the equations**:
- We now have two equations:
\[ A + B = 60^\circ \quad (1) \]
\[ A - B = 0^\circ \quad (2) \]

3. **Solve for \( A \) and \( B \)**:
- To find \( A \) and \( B \), we can add equations (1) and (2):
\[ (A + B) + (A - B) = 60^\circ + 0^\circ \]
\[ 2A = 60^\circ \]
\[ A = 30^\circ \]
- Now, substitute \( A = 30^\circ \) back into equation (1) to find \( B \):
\[ 30^\circ + B = 60^\circ \]
\[ B = 60^\circ - 30^\circ = 30^\circ \]

4. **Calculate \( \sin(A + B) \)**:
- Now that we have \( A + B = 60^\circ \):
\[ \sin(A + B) = \sin(60^\circ) = \frac{\sqrt{3}}{2} \]

5. **Calculate \( \cos(A - B) \)**:
- Now that we have \( A - B = 0^\circ \):
\[ \cos(A - B) = \cos(0^\circ) = 1 \]

### Final Answers:
- \( \sin(A + B) = \frac{\sqrt{3}}{2} \)
- \( \cos(A - B) = 1 \)