The set of all poddible ualuse of `alpha` in `[-pi,pi]` such that `sqrt((1-sinalpha)/(a+sinalpha))` is equal to sec `alpha-tanalpha,` is

Options

A. `[0,pi//2)`
B. `[0,pi//2)uu(pi//2,pi)`
C. `[-pi,0]`
D. `(-pi//2,pi//2)

Explanation

To solve the equation \( \sqrt{\frac{1 - \sin \alpha}{1 + \sin \alpha}} = \sec \alpha - \tan \alpha \), we will follow these steps:

### Step 1: Simplify the left-hand side
We start with the left-hand side:
\[ \sqrt{\frac{1 - \sin \alpha}{1 + \sin \alpha}} \]

### Step 2: Rationalize the expression
To simplify, we can multiply the numerator and denominator by \( 1 - \sin \alpha \):
\[ \sqrt{\frac{(1 - \sin \alpha)(1 - \sin \alpha)}{(1 + \sin \alpha)(1 - \sin \alpha)}} \]
This gives us:
\[ \sqrt{\frac{(1 - \sin \alpha)^2}{1 - \sin^2 \alpha}} = \sqrt{\frac{(1 - \sin \alpha)^2}{\cos^2 \alpha}} \]
Thus, we can simplify it to:
\[ \frac{1 - \sin \alpha}{\cos \alpha} \]

### Step 3: Write the right-hand side
Now, we rewrite the right-hand side:
\[ \sec \alpha - \tan \alpha = \frac{1}{\cos \alpha} - \frac{\sin \alpha}{\cos \alpha} = \frac{1 - \sin \alpha}{\cos \alpha} \]

### Step 4: Set the two sides equal
Now we have:
\[ \frac{1 - \sin \alpha}{\cos \alpha} = \frac{1 - \sin \alpha}{\cos \alpha} \]
This equation holds true for all values of \( \alpha \) where \( \cos \alpha \neq 0 \) and \( 1 - \sin \alpha \neq 0 \).

### Step 5: Determine the restrictions
1. \( \cos \alpha \neq 0 \) implies \( \alpha \neq \frac{\pi}{2} + k\pi \) for any integer \( k \).
2. \( 1 - \sin \alpha \neq 0 \) implies \( \sin \alpha \neq 1 \) which gives \( \alpha \neq \frac{\pi}{2} \).

### Step 6: Find the solution set
The possible values of \( \alpha \) in the interval \( [-\pi, \pi] \) excluding \( \frac{\pi}{2} \) are:
\[ [-\pi, -\frac{\pi}{2}) \cup (-\frac{\pi}{2}, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi] \]

### Final Answer
Thus, the set of all possible values of \( \alpha \) in \( [-\pi, \pi] \) such that the equation holds is:
\[ [-\pi, -\frac{\pi}{2}) \cup (-\frac{\pi}{2}, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi] \]