If `(2sinA)/(1+sinA+cosA)=k` then `(1+sinA-cosA)/(1+sinA)=`

Options

A. `k/2`
B. k
C. 2k
D. `1/k`

Answer

To solve the problem, we start with the given equation:

\[ \frac{2 \sin A}{1 + \sin A + \cos A} = k \]

We need to find the value of

\[ \frac{1 + \sin A - \cos A}{1 + \sin A} \]

### Step 1: Rationalize the expression

We can rationalize the expression by multiplying the numerator and denominator of the left-hand side by \(1 + \sin A - \cos A\):

\[ \frac{2 \sin A}{1 + \sin A + \cos A} \cdot \frac{1 + \sin A - \cos A}{1 + \sin A - \cos A} \]

### Step 2: Simplify the numerator

The numerator becomes:

\[ 2 \sin A (1 + \sin A - \cos A) = 2 \sin A + 2 \sin^2 A - 2 \sin A \cos A \]

### Step 3: Simplify the denominator

The denominator can be simplified using the difference of squares:

\[ (1 + \sin A + \cos A)(1 + \sin A - \cos A) = (1 + \sin A)^2 - \cos^2 A \]

Using the identity \(\cos^2 A = 1 - \sin^2 A\), we can rewrite the denominator:

\[ (1 + \sin A)^2 - (1 - \sin^2 A) = (1 + \sin A)^2 - 1 + \sin^2 A \]

Expanding \((1 + \sin A)^2\):

\[ 1 + 2\sin A + \sin^2 A - 1 + \sin^2 A = 2\sin^2 A + 2\sin A \]

### Step 4: Combine the results

Now we have:

\[ \frac{2 \sin A + 2 \sin^2 A - 2 \sin A \cos A}{2 \sin^2 A + 2 \sin A} \]

### Step 5: Factor out common terms

We can factor out \(2\) from both the numerator and denominator:

\[ \frac{2(\sin A + \sin^2 A - \sin A \cos A)}{2(\sin^2 A + \sin A)} = \frac{\sin A + \sin^2 A - \sin A \cos A}{\sin^2 A + \sin A} \]

### Step 6: Substitute back to find k

From the original equation, we know:

\[ \frac{2 \sin A}{1 + \sin A + \cos A} = k \]

Thus, we can express the result in terms of \(k\):

\[ \frac{1 + \sin A - \cos A}{1 + \sin A} = k \]

### Final Answer:

Thus, the value of \(\frac{1 + \sin A - \cos A}{1 + \sin A}\) is:

\[ \boxed{k} \]