If `alpha and beta` are complementary angles, then what is `sqrt(cos alpha cosec beta -cosalpha sinbeta)` equal to ?

Options

A. `sec beta`
B. `cos alpha`
C. `sin alpha`
D. `-tan beta`

Answer

To solve the problem, we need to find the value of \( \sqrt{\cos \alpha \csc \beta - \cos \alpha \sin \beta} \) given that \( \alpha \) and \( \beta \) are complementary angles. This means that \( \alpha + \beta = 90^\circ \) or \( \beta = 90^\circ - \alpha \).

### Step-by-Step Solution:

1. **Understanding Complementary Angles**:
Since \( \alpha \) and \( \beta \) are complementary, we can use the identities:
\[ \csc \beta = \csc(90^\circ - \alpha) = \sec \alpha \]
and
\[ \sin \beta = \sin(90^\circ - \alpha) = \cos \alpha. \]

2. **Substituting Values**:
Now, we substitute \( \csc \beta \) and \( \sin \beta \) into the expression:
\[ \sqrt{\cos \alpha \csc \beta - \cos \alpha \sin \beta} = \sqrt{\cos \alpha \sec \alpha - \cos \alpha \cos \alpha}. \]

3. **Simplifying the Expression**:
We know that \( \cos \alpha \sec \alpha = 1 \) (since \( \sec \alpha = \frac{1}{\cos \alpha} \)):
\[ = \sqrt{1 - \cos^2 \alpha}. \]

4. **Using the Pythagorean Identity**:
From the Pythagorean identity, we have:
\[ 1 - \cos^2 \alpha = \sin^2 \alpha. \]
Therefore, we can rewrite the expression as:
\[ = \sqrt{\sin^2 \alpha}. \]

5. **Taking the Square Root**:
The square root of \( \sin^2 \alpha \) is:
\[ = |\sin \alpha|. \]
Since \( \alpha \) is an angle, we can assume \( \sin \alpha \) is non-negative in the range of \( 0^\circ \) to \( 90^\circ \):
\[ = \sin \alpha. \]

### Final Answer:
Thus, the value of \( \sqrt{\cos \alpha \csc \beta - \cos \alpha \sin \beta} \) is \( \sin \alpha \).