`0 lt A lt 90^@ : 0 lt B lt 90^@ , sin A=12/13` and cos B=`4/5`. Find the vaue of sin (A+B).

Answer

To solve the problem, we need to find the value of \( \sin(A + B) \) given that \( \sin A = \frac{12}{13} \) and \( \cos B = \frac{4}{5} \).

### Step-by-Step Solution:

1. **Use the sine addition formula:**
\[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]

2. **Identify known values:**
- We know \( \sin A = \frac{12}{13} \) and \( \cos B = \frac{4}{5} \).
- We need to find \( \cos A \) and \( \sin B \).

3. **Find \( \cos A \):**
- We can use the Pythagorean identity:
\[ \sin^2 A + \cos^2 A = 1 \]
- Substitute \( \sin A \):
\[ \left(\frac{12}{13}\right)^2 + \cos^2 A = 1 \]
- Calculate \( \left(\frac{12}{13}\right)^2 \):
\[ \frac{144}{169} + \cos^2 A = 1 \]
- Rearranging gives:
\[ \cos^2 A = 1 - \frac{144}{169} = \frac{169 - 144}{169} = \frac{25}{169} \]
- Taking the square root:
\[ \cos A = \sqrt{\frac{25}{169}} = \frac{5}{13} \]

4. **Find \( \sin B \):**
- Again, use the Pythagorean identity:
\[ \sin^2 B + \cos^2 B = 1 \]
- Substitute \( \cos B \):
\[ \sin^2 B + \left(\frac{4}{5}\right)^2 = 1 \]
- Calculate \( \left(\frac{4}{5}\right)^2 \):
\[ \sin^2 B + \frac{16}{25} = 1 \]
- Rearranging gives:
\[ \sin^2 B = 1 - \frac{16}{25} = \frac{25 - 16}{25} = \frac{9}{25} \]
- Taking the square root:
\[ \sin B = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

5. **Substitute values into the sine addition formula:**
\[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]
\[ = \left(\frac{12}{13}\right)\left(\frac{4}{5}\right) + \left(\frac{5}{13}\right)\left(\frac{3}{5}\right) \]

6. **Calculate each term:**
- First term:
\[ \frac{12 \times 4}{13 \times 5} = \frac{48}{65} \]
- Second term:
\[ \frac{5 \times 3}{13 \times 5} = \frac{15}{65} \]

7. **Combine the results:**
\[ \sin(A + B) = \frac{48}{65} + \frac{15}{65} = \frac{63}{65} \]

### Final Answer:
\[ \sin(A + B) = \frac{63}{65} \]