In ` Delta ABC , angle B = 90^(@) ` find the values of :
` sin A cos C + cos A sin C `

Answer

To solve the problem, we need to find the value of \( \sin A \cos C + \cos A \sin C \) in triangle \( ABC \) where \( \angle B = 90^\circ \).

### Step-by-Step Solution:

1. **Identify the Angles**: In triangle \( ABC \), we know that \( \angle B = 90^\circ \). Therefore, the sum of the angles in a triangle gives us:
\[ A + B + C = 180^\circ \]
Since \( B = 90^\circ \), we can substitute:
\[ A + 90^\circ + C = 180^\circ \]

2. **Simplify the Equation**: Rearranging the equation gives us:
\[ A + C = 180^\circ - 90^\circ = 90^\circ \]
This means that \( A + C = 90^\circ \).

3. **Use the Trigonometric Identity**: The expression we need to evaluate is:
\[ \sin A \cos C + \cos A \sin C \]
This expression is recognized as the sine addition formula:
\[ \sin(A + C) \]

4. **Substitute the Value of \( A + C \)**: Since we found that \( A + C = 90^\circ \), we can substitute this into the sine function:
\[ \sin(A + C) = \sin(90^\circ) \]

5. **Evaluate \( \sin(90^\circ) \)**: We know that:
\[ \sin(90^\circ) = 1 \]

6. **Final Answer**: Therefore, the value of \( \sin A \cos C + \cos A \sin C \) is:
\[ \boxed{1} \]