Let ` a= ( 2 sin x )/(1 + sin x + cos x ) and b = ( c )/(1+ sin x ) . ` Then a = b , if c= ?

Options

A. ` 1 + sin x cos x `
B. ` 1 + sin x - cos x `
C. ` 1 + cos x - sin x `
D. ` 1- sin x cos x `

Answer

To solve the problem, we need to find the value of \( c \) such that \( a = b \) where:

\[ a = \frac{2 \sin x}{1 + \sin x + \cos x} \]
\[ b = \frac{c}{1 + \sin x} \]

### Step 1: Set the equations equal to each other
Since \( a = b \), we can write:

\[ \frac{2 \sin x}{1 + \sin x + \cos x} = \frac{c}{1 + \sin x} \]

### Step 2: Cross-multiply to eliminate the fractions
Cross-multiplying gives us:

\[ 2 \sin x (1 + \sin x) = c (1 + \sin x + \cos x) \]

### Step 3: Expand both sides
Expanding both sides results in:

\[ 2 \sin x + 2 \sin^2 x = c + c \sin x + c \cos x \]

### Step 4: Rearranging the equation
Rearranging the equation to isolate \( c \):

\[ 2 \sin^2 x + 2 \sin x - c \sin x - c \cos x - c = 0 \]

### Step 5: Collect like terms
Rearranging gives us:

\[ (2 - c) \sin x + 2 \sin^2 x - c \cos x - c = 0 \]

### Step 6: Compare coefficients
For the equation to hold for all \( x \), the coefficients of \( \sin x \) and \( \cos x \) must be equal. Thus, we can derive two equations:

1. \( 2 - c = 0 \) (coefficient of \( \sin x \))
2. \( -c = 0 \) (coefficient of \( \cos x \))

### Step 7: Solve for \( c \)
From the first equation:

\[ c = 2 \]

From the second equation, since it is \( -c = 0 \), it confirms that \( c \) must be \( 2 \).

### Final Answer
Thus, the value of \( c \) is:

\[ \boxed{2} \]