The set of all possible values of `alpha` in `[-pi,pi]` such that `sqrt((1-sinalpha)/(1+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: Start with the given equation
We have:
\[ \sqrt{\frac{1 - \sin \alpha}{1 + \sin \alpha}} = \sec \alpha - \tan \alpha \]

### Step 2: Rationalize the left-hand side
To simplify the left-hand side, we can multiply the numerator and denominator by \( 1 - \sin \alpha \):
\[ \sqrt{\frac{(1 - \sin \alpha)(1 - \sin \alpha)}{(1 + \sin \alpha)(1 - \sin \alpha)}} = \sqrt{\frac{(1 - \sin \alpha)^2}{1 - \sin^2 \alpha}} \]

### Step 3: Simplify the expression
Recall that \( 1 - \sin^2 \alpha = \cos^2 \alpha \):
\[ \sqrt{\frac{(1 - \sin \alpha)^2}{\cos^2 \alpha}} = \frac{1 - \sin \alpha}{|\cos \alpha|} \]

### Step 4: Consider the sign of \( \cos \alpha \)
Since we want the expression to match \( \sec \alpha - \tan \alpha \), we will consider the case where \( \cos \alpha > 0 \) (which occurs in the first and fourth quadrants):
\[ \frac{1 - \sin \alpha}{\cos \alpha} = \sec \alpha - \tan \alpha \]

### Step 5: Rewrite the right-hand side
We know that:
\[ \sec \alpha = \frac{1}{\cos \alpha}, \quad \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \]
Thus,
\[ \sec \alpha - \tan \alpha = \frac{1 - \sin \alpha}{\cos \alpha} \]

### Step 6: Set the two sides equal
Now we have:
\[ \frac{1 - \sin \alpha}{\cos \alpha} = \frac{1 - \sin \alpha}{\cos \alpha} \]
This is true as long as \( 1 - \sin \alpha \neq 0 \).

### Step 7: Solve for \( \alpha \)
The condition \( 1 - \sin \alpha = 0 \) gives us \( \sin \alpha = 1 \), which occurs at \( \alpha = \frac{\pi}{2} \). However, we need to check the intervals where \( \cos \alpha > 0 \).

### Step 8: Identify the intervals
Since \( \cos \alpha > 0 \) in the intervals:
\[ -\frac{\pi}{2} < \alpha < \frac{\pi}{2} \]

### Conclusion
Thus, the set of all possible values of \( \alpha \) in the interval \( [-\pi, \pi] \) such that the original equation holds is:
\[ \alpha \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \]