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`

Options

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

Explanation

To solve the problem, we start with the given expression for \( x \):

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

We need to find the value of:

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

### Step 1: Multiply numerator and denominator by \( 1 - \cos \theta + \sin \theta \)

We can rewrite \( x \) by multiplying both the numerator and denominator 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 2: Simplify the denominator using the identity \( (A + B)(A - B) = A^2 - B^2 \)

Let \( A = 1 + \sin \theta \) and \( B = \cos \theta \). Thus, we can apply the identity:

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

### Step 3: Expand the denominator

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

\[ (1 + \sin \theta)^2 = 1 + 2 \sin \theta + \sin^2 \theta \]

Using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \):

\[ \cos^2 \theta = 1 - \sin^2 \theta \]

Thus, the denominator becomes:

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

### Step 4: Substitute back into the expression for \( x \)

Now substituting back into the expression for \( x \):

\[ x = \frac{2 \sin \theta (1 - \cos \theta + \sin \theta)}{2 (\sin^2 \theta + \sin \theta)} \]

### Step 5: Cancel out common factors

We can cancel \( 2 \sin \theta \) from the numerator and denominator:

\[ x = \frac{1 - \cos \theta + \sin \theta}{1 + \sin \theta} \]

### Conclusion

Thus, we find that:

\[ \frac{1 - \cos \theta + \sin \theta}{1 + \sin \theta} = x \]

So the answer is:

**Option (c) \( x \)**