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

Options

A. `2npi+(pi)/(4)and 2npi+(3pi)/(4)`
B. `2npi+(pi)/(4)and 2npi+(pi)/(4)`
C. `2npi+(3pi)/(4)and 2npi+(pi)/(4)`
D. `-ooand +oo`

Answer

To solve the equation \( \frac{2 \cos(A)}{2} = \sqrt{1 + \sin A} + \sqrt{1 - \sin A} \), we will follow these steps:

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

### Step 2: Simplify the Right Side
Using the identity \( \sqrt{1 + \sin A} + \sqrt{1 - \sin A} \), we can rewrite it as:
\[ \sqrt{(1 + \sin A)(1 - \sin A)} + \sqrt{1 - \sin^2 A} \]
Since \( 1 - \sin^2 A = \cos^2 A \), we have:
\[ \sqrt{1 - \sin^2 A} = \cos A \]
Thus, the right side becomes:
\[ \sqrt{(1 + \sin A)(1 - \sin A)} + \cos A \]

### Step 3: Use the Identity for Difference of Squares
The expression \( (1 + \sin A)(1 - \sin A) \) can be simplified to:
\[ 1 - \sin^2 A = \cos^2 A \]
Thus, we have:
\[ \sqrt{\cos^2 A} + \cos A \]
This simplifies to:
\[ |\cos A| + \cos A \]

### Step 4: Analyze Cases for \( |\cos A| \)
1. If \( \cos A \geq 0 \):
\[ |\cos A| = \cos A \implies |\cos A| + \cos A = 2\cos A \]
Therefore, the equation becomes:
\[ \cos A = 2\cos A \]
This leads to:
\[ \cos A = 0 \]
Which implies:
\[ A = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]

2. If \( \cos A < 0 \):
\[ |\cos A| = -\cos A \implies |\cos A| + \cos A = 0 \]
Therefore, the equation becomes:
\[ \cos A = 0 \]
Which again leads to:
\[ A = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]

### Step 5: Find the Range for \( A/2 \)
Since \( A = \frac{\pi}{2} + n\pi \), we can find \( A/2 \):
\[ \frac{A}{2} = \frac{\pi}{4} + \frac{n\pi}{2} \]

### Step 6: Determine the Range
For \( n = 0 \):
\[ \frac{A}{2} = \frac{\pi}{4} \]
For \( n = 1 \):
\[ \frac{A}{2} = \frac{3\pi}{4} \]
For \( n = -1 \):
\[ \frac{A}{2} = -\frac{\pi}{4} \]
Thus, \( \frac{A}{2} \) lies in the range:
\[ \left[-\frac{\pi}{4}, \frac{3\pi}{4}\right] \]

### Final Answer
The value of \( \frac{A}{2} \) lies between \( -\frac{\pi}{4} \) and \( \frac{3\pi}{4} \).

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