`sqrt(1+sinA)=-(sin(A/2)+cos (A/2))` is true if :

Options

A. A. `(3pi)/(2) lt A lt (5pi)/(2)`
B. B. `(pi)/(2) lt A lt (3pi)/(2)`
C. C. `(3pi)/(2) lt A lt(7pi)/(2)`
D. D. `0 lt A lt (3 pi)/(2)`

Answer

To solve the equation \( \sqrt{1 + \sin A} = -(\sin(A/2) + \cos(A/2)) \), we need to analyze both sides of the equation step by step.

### Step 1: Analyze the Left-Hand Side (LHS)
The left-hand side is given as:
\[ LHS = \sqrt{1 + \sin A} \]
Using the identity \( 1 + \sin A = \sin^2(A/2) + \cos^2(A/2) + 2\sin(A/2)\cos(A/2) \), we can rewrite it as:
\[ LHS = \sqrt{(\sin(A/2) + \cos(A/2))^2} \]
Thus, we have:
\[ LHS = |\sin(A/2) + \cos(A/2)| \]

### Step 2: Analyze the Right-Hand Side (RHS)
The right-hand side is given as:
\[ RHS = -(\sin(A/2) + \cos(A/2)) \]
This means that the RHS is negative.

### Step 3: Set Up the Condition
For the equation \( |\sin(A/2) + \cos(A/2)| = -(\sin(A/2) + \cos(A/2)) \) to hold true, the expression inside the absolute value must be negative:
\[ \sin(A/2) + \cos(A/2) < 0 \]

### Step 4: Determine When \( \sin(A/2) + \cos(A/2) < 0 \)
To find when \( \sin(A/2) + \cos(A/2) < 0 \), we can analyze the function:
\[ \sin(A/2) + \cos(A/2) = 0 \]
This occurs when:
\[ \tan(A/2) = -1 \]
The general solutions for this equation are:
\[ A/2 = \frac{3\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \]
Thus, multiplying by 2 gives:
\[ A = \frac{3\pi}{2} + 2n\pi \]

### Step 5: Find the Range of A
The expression \( \sin(A/2) + \cos(A/2) < 0 \) occurs in the intervals where \( A/2 \) is in the third and fourth quadrants. The specific intervals for \( A/2 \) are:
\[ \frac{3\pi}{4} < A/2 < \frac{7\pi}{4} \]
Multiplying through by 2 gives:
\[ \frac{3\pi}{2} < A < \frac{7\pi}{2} \]

### Conclusion
The condition for the equation \( \sqrt{1 + \sin A} = -(\sin(A/2) + \cos(A/2)) \) to hold true is:
\[ A \in \left( \frac{3\pi}{2}, \frac{7\pi}{2} \right) \]