If `A+B=90^(@)`, then what is `sqrt(sinAsecB-sinAcosB)`equal to ?

Options

A. sin A
B. cosA
C. tanA
D. 0

Explanation

To solve the problem, we need to find the expression \( \sqrt{\sin A \sec B - \sin A \cos B} \) given that \( A + B = 90^\circ \).

### Step-by-Step Solution:

1. **Use the Identity**: Since \( A + B = 90^\circ \), we can express \( B \) as \( B = 90^\circ - A \).

2. **Substitute for \( \sec B \)**: Recall that \( \sec B = \frac{1}{\cos B} \). Using the identity for \( B \):
\[ \sec B = \sec(90^\circ - A) = \csc A \]

3. **Substitute into the Expression**: Now substitute \( \sec B \) in the original expression:
\[ \sqrt{\sin A \sec B - \sin A \cos B} = \sqrt{\sin A \csc A - \sin A \cos(90^\circ - A)} \]

4. **Simplify \( \cos(90^\circ - A) \)**: We know that \( \cos(90^\circ - A) = \sin A \). Thus:
\[ \sqrt{\sin A \csc A - \sin A \sin A} \]

5. **Rewrite \( \csc A \)**: Since \( \csc A = \frac{1}{\sin A} \), we can rewrite:
\[ \sqrt{\sin A \cdot \frac{1}{\sin A} - \sin^2 A} = \sqrt{1 - \sin^2 A} \]

6. **Use the Pythagorean Identity**: Recall that \( 1 - \sin^2 A = \cos^2 A \):
\[ \sqrt{1 - \sin^2 A} = \sqrt{\cos^2 A} \]

7. **Final Simplification**: The square root of \( \cos^2 A \) is \( |\cos A| \). Since \( A \) is an angle in a triangle, we can assume \( \cos A \) is non-negative:
\[ \sqrt{\cos^2 A} = \cos A \]

### Final Answer:
Thus, the expression \( \sqrt{\sin A \sec B - \sin A \cos B} \) simplifies to:
\[ \cos A \]