If `2cos""(A)/(2)=sqrt(1+sinA)+sqrt(1-sinA), thenA/2` iles between,

Answer

To solve the equation \( \frac{2 \cos(A/2)}{2} = \sqrt{1 + \sin A} + \sqrt{1 - \sin A} \), we will simplify and analyze the expressions step by step.

### Step 1: Simplify the Left Side
The left side simplifies to:
\[ \cos(A/2) \]

### Step 2: Simplify the Right Side
The right side is:
\[ \sqrt{1 + \sin A} + \sqrt{1 - \sin A} \]
Using the identity \( \sin A = 2 \sin(A/2) \cos(A/2) \) and \( 1 = \sin^2(A/2) + \cos^2(A/2) \), we can rewrite \( 1 + \sin A \) and \( 1 - \sin A \).

### Step 3: Rewrite \( \sqrt{1 + \sin A} \)
We can express \( 1 + \sin A \) as:
\[ 1 + \sin A = 1 + 2 \sin(A/2) \cos(A/2) = \sin^2(A/2) + \cos^2(A/2) + 2 \sin(A/2) \cos(A/2) \]
This can be rewritten as:
\[ (\sin(A/2) + \cos(A/2))^2 \]
Thus,
\[ \sqrt{1 + \sin A} = |\sin(A/2) + \cos(A/2)| \]

### Step 4: Rewrite \( \sqrt{1 - \sin A} \)
Similarly, for \( 1 - \sin A \):
\[ 1 - \sin A = 1 - 2 \sin(A/2) \cos(A/2) = \sin^2(A/2) + \cos^2(A/2) - 2 \sin(A/2) \cos(A/2) \]
This can be rewritten as:
\[ (\sin(A/2) - \cos(A/2))^2 \]
Thus,
\[ \sqrt{1 - \sin A} = |\sin(A/2) - \cos(A/2)| \]

### Step 5: Combine the Right Side
Now, we can express the right side as:
\[ |\sin(A/2) + \cos(A/2)| + |\sin(A/2) - \cos(A/2)| \]

### Step 6: Set Up the Equation
Now we have:
\[ \cos(A/2) = |\sin(A/2) + \cos(A/2)| + |\sin(A/2) - \cos(A/2)| \]

### Step 7: Analyze Cases
We need to analyze the cases based on the signs of \( \sin(A/2) \) and \( \cos(A/2) \).

1. **Case 1**: If \( \sin(A/2) \geq \cos(A/2) \):
\[ \cos(A/2) = 2 \sin(A/2) \]
This leads to \( \tan(A/2) = \frac{1}{2} \).

2. **Case 2**: If \( \sin(A/2) < \cos(A/2) \):
\[ \cos(A/2) = 0 \]
This leads to \( A/2 = \frac{\pi}{2} + n\pi \).

### Step 8: Find the Range of \( A/2 \)
From the first case, \( \tan(A/2) = \frac{1}{2} \) implies:
\[ A/2 = \tan^{-1}\left(\frac{1}{2}\right) \]
This value lies between \( -\frac{\pi}{4} \) and \( \frac{\pi}{4} \).

### Conclusion
Thus, the value of \( A/2 \) lies between:
\[ -\frac{\pi}{4} < A/2 < \frac{\pi}{4} \]