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

Answer

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

### Step-by-Step Solution:

1. **Understand the Triangle Properties**:
In triangle \( ABC \), since \( \angle B = 90^\circ \), we know that the sum of the angles in a triangle is \( 180^\circ \). Therefore, we can write:
\[ A + B + C = 180^\circ \]
Substituting \( B = 90^\circ \):
\[ A + 90^\circ + C = 180^\circ \]
Simplifying this gives:
\[ A + C = 90^\circ \]

2. **Use the Cosine Addition Formula**:
We need to evaluate \( \cos A \cos C - \sin A \sin C \). This expression can be rewritten using the cosine of the sum of angles:
\[ \cos A \cos C - \sin A \sin C = \cos(A + C) \]

3. **Substitute the Value of \( A + C \)**:
From step 1, we found that \( A + C = 90^\circ \). Therefore, we can substitute this into our expression:
\[ \cos(A + C) = \cos(90^\circ) \]

4. **Evaluate \( \cos(90^\circ) \)**:
The value of \( \cos(90^\circ) \) is:
\[ \cos(90^\circ) = 0 \]

5. **Conclusion**:
Thus, we find that:
\[ \cos A \cos C - \sin A \sin C = 0 \]

### Final Answer:
The value of \( \cos A \cos C - \sin A \sin C \) is \( 0 \).