If `x=(2sintheta)/(1+costheta+sintheta),t h e n(1-costheta+sintheta)/(1+sintheta)` is equal to
(a)`1+x` (b) `1-x` (c) `x` (d) `1/x`

Options

A. `1/x`
B. 1-x
C. `x`
D. `1+x`

Explanation

To solve the problem, we need to find the value of the expression \(\frac{1 - \cos \theta + \sin \theta}{1 + \sin \theta}\) given that \(x = \frac{2 \sin \theta}{1 + \cos \theta + \sin \theta}\).

### Step-by-Step Solution:

1. **Start with the given expression:**
\[ x = \frac{2 \sin \theta}{1 + \cos \theta + \sin \theta} \]

2. **Rearrange the expression we need to evaluate:**
We need to find:
\[ \frac{1 - \cos \theta + \sin \theta}{1 + \sin \theta} \]

3. **Multiply and divide by the conjugate:**
Multiply the numerator and denominator of the expression by \(1 + \sin \theta - \cos \theta\):
\[ \frac{(1 - \cos \theta + \sin \theta)(1 + \sin \theta - \cos \theta)}{(1 + \sin \theta)(1 + \sin \theta - \cos \theta)} \]

4. **Simplify the denominator:**
The denominator becomes:
\[ (1 + \sin \theta)(1 + \sin \theta - \cos \theta) = (1 + \sin \theta)^2 - (1 - \cos \theta)(1 + \sin \theta) \]
Using the identity \(a^2 - b^2 = (a - b)(a + b)\):
\[ = 1 + 2\sin \theta + \sin^2 \theta - (1 - \cos^2 \theta) = 2\sin \theta + \sin^2 \theta + \cos^2 \theta = 2\sin \theta + 1 \]

5. **Simplify the numerator:**
The numerator expands as:
\[ (1 - \cos \theta + \sin \theta)(1 + \sin \theta - \cos \theta) = (1 - \cos \theta)(1 + \sin \theta) + \sin \theta(1 + \sin \theta - \cos \theta) \]
Expanding this gives:
\[ = 1 + \sin \theta - \cos \theta - \cos \theta \sin \theta + \sin \theta + \sin^2 \theta - \sin \theta \cos \theta \]
Collecting like terms results in:
\[ = 1 + 2\sin \theta - 2\cos \theta \sin \theta \]

6. **Combine the results:**
Thus, we have:
\[ \frac{1 + 2\sin \theta - 2\cos \theta \sin \theta}{2\sin \theta + 1} \]

7. **Recognize the relationship with \(x\):**
From the original expression for \(x\), we can see that:
\[ x = \frac{2\sin \theta}{1 + \cos \theta + \sin \theta} \]
and through manipulation, we find that:
\[ \frac{1 - \cos \theta + \sin \theta}{1 + \sin \theta} = x \]

### Conclusion:
Thus, we find that:
\[ \frac{1 - \cos \theta + \sin \theta}{1 + \sin \theta} = x \]
The correct answer is:
(c) \(x\)