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

Answer

To solve the problem, we need to verify the expression given in the question and find out which option it corresponds to.

Given:
\[ x = \frac{2 \sin \theta}{1 + \cos \theta + \sin \theta} \]

We need to find:
\[ \frac{1 - \cos \theta + \sin \theta}{1 + \sin \theta} \]

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

**Step 2: Multiply and divide by the conjugate.**
To simplify the expression, we multiply and divide by \(1 - \cos \theta + \sin \theta\):
\[ x = \frac{2 \sin \theta (1 - \cos \theta + \sin \theta)}{(1 + \cos \theta + \sin \theta)(1 - \cos \theta + \sin \theta)} \]

**Step 3: Expand the denominator.**
Using the difference of squares:
\[ (1 + \cos \theta + \sin \theta)(1 - \cos \theta + \sin \theta) = (1 + \sin \theta)^2 - \cos^2 \theta \]
This simplifies to:
\[ 1 + 2\sin \theta + \sin^2 \theta - \cos^2 \theta \]
Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\), we can replace \(-\cos^2 \theta\) with \(-1 + \sin^2 \theta\):
\[ = 1 + 2\sin \theta + \sin^2 \theta - (1 - \sin^2 \theta) = 2\sin^2 \theta + 2\sin \theta \]

**Step 4: Substitute back into x.**
Now substituting this back into the expression for \(x\):
\[ x = \frac{2 \sin \theta (1 - \cos \theta + \sin \theta)}{2(\sin^2 \theta + \sin \theta)} \]
We can cancel \(2\) from numerator and denominator:
\[ x = \frac{\sin \theta (1 - \cos \theta + \sin \theta)}{\sin^2 \theta + \sin \theta} \]

**Step 5: Simplify further.**
Now we can simplify this expression:
\[ = \frac{1 - \cos \theta + \sin \theta}{1 + \sin \theta} \]

**Conclusion:**
Thus, we have shown that:
\[ \frac{1 - \cos \theta + \sin \theta}{1 + \sin \theta} = x \]
So the answer is:
**(c) \(x\)**