If `alpha in [(pi)/(2),pi]` then the value of
`sqrt(1+ sin alpha)-sqrt(1 - sin alpha)` is equal to

Options

A. `cos theta`
B. `sin theta`
C. `2 cos theta`
D. `2 sin theta`

Answer

To solve the problem, we need to find the value of the expression \( \sqrt{1 + \sin \alpha} - \sqrt{1 - \sin \alpha} \) given that \( \alpha \) is in the interval \( \left[\frac{\pi}{2}, \pi\right] \).

### Step-by-Step Solution:

1. **Start with the expression:**
\[ \sqrt{1 + \sin \alpha} - \sqrt{1 - \sin \alpha} \]

2. **Rationalize the expression:**
To simplify this, we can multiply and divide by the conjugate:
\[ \frac{(\sqrt{1 + \sin \alpha} - \sqrt{1 - \sin \alpha})(\sqrt{1 + \sin \alpha} + \sqrt{1 - \sin \alpha})}{\sqrt{1 + \sin \alpha} + \sqrt{1 - \sin \alpha}} \]
This gives us:
\[ \frac{(1 + \sin \alpha) - (1 - \sin \alpha)}{\sqrt{1 + \sin \alpha} + \sqrt{1 - \sin \alpha}} \]

3. **Simplify the numerator:**
The numerator simplifies to:
\[ 1 + \sin \alpha - 1 + \sin \alpha = 2 \sin \alpha \]
So, we have:
\[ \frac{2 \sin \alpha}{\sqrt{1 + \sin \alpha} + \sqrt{1 - \sin \alpha}} \]

4. **Substituting values for \( \alpha \):**
Since \( \alpha \) is in the interval \( \left[\frac{\pi}{2}, \pi\right] \), we know that \( \sin \alpha \) will be in the range \( [1, 0] \).

5. **Evaluate the denominator:**
- When \( \alpha = \frac{\pi}{2} \), \( \sin \alpha = 1 \):
\[ \sqrt{1 + \sin \alpha} = \sqrt{2}, \quad \sqrt{1 - \sin \alpha} = 0 \]
Thus, the denominator becomes \( \sqrt{2} + 0 = \sqrt{2} \).

- When \( \alpha = \pi \), \( \sin \alpha = 0 \):
\[ \sqrt{1 + \sin \alpha} = 1, \quad \sqrt{1 - \sin \alpha} = 1 \]
Thus, the denominator becomes \( 1 + 1 = 2 \).

6. **Final expression:**
Therefore, the expression simplifies to:
\[ \frac{2 \sin \alpha}{\sqrt{1 + \sin \alpha} + \sqrt{1 - \sin \alpha}} \]
For \( \alpha = \frac{\pi}{2} \):
\[ \frac{2 \cdot 1}{\sqrt{2}} = \sqrt{2} \]
For \( \alpha = \pi \):
\[ \frac{2 \cdot 0}{2} = 0 \]

7. **Conclusion:**
The value of \( \sqrt{1 + \sin \alpha} - \sqrt{1 - \sin \alpha} \) varies between \( \sqrt{2} \) and \( 0 \) as \( \alpha \) varies from \( \frac{\pi}{2} \) to \( \pi \).

### Final Answer:
The value of \( \sqrt{1 + \sin \alpha} - \sqrt{1 - \sin \alpha} \) is \( 2 \cos \left(\frac{\alpha}{2}\right) \).