In `DeltaABC`, right angled at B, find the value of sin A cos C + cosA sinC.

Answer

To solve the problem, we need to find the value of \( \sin A \cos C + \cos A \sin C \) in triangle \( \Delta ABC \) where angle \( B \) is the right angle.

### Step-by-Step Solution:

1. **Identify the Angles**: In triangle \( ABC \), since it is right-angled at \( B \), we know that \( A + B + C = 180^\circ \). Given that \( B = 90^\circ \), we can deduce that:
\[ A + C = 90^\circ \]

2. **Use the Sine Addition Formula**: The expression \( \sin A \cos C + \cos A \sin C \) can be recognized as the sine of the sum of angles:
\[ \sin A \cos C + \cos A \sin C = \sin(A + C) \]

3. **Substitute the Value of \( A + C \)**: Since we established that \( A + C = 90^\circ \), we can substitute this into our equation:
\[ \sin(A + C) = \sin(90^\circ) \]

4. **Calculate \( \sin(90^\circ) \)**: The value of \( \sin(90^\circ) \) is:
\[ \sin(90^\circ) = 1 \]

5. **Final Result**: Therefore, the value of \( \sin A \cos C + \cos A \sin C \) is:
\[ \sin A \cos C + \cos A \sin C = 1 \]

### Conclusion:
The final answer is:
\[ \sin A \cos C + \cos A \sin C = 1 \]