Statement:1 If `theta` is acute and `1+ costheta = k`, then `sin (theta/2)` is `sqrt((2-k)/2)`. and Statement-2: `2sin^(2)(x/2)=1-cosx`.

Options

A. Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation for statement-1
B. Statement-1 is true, Statement-2 is true, Statement-2 is not a correct expanation for statement-1
C. Statement-1 is True, Statement-2 is false
D. Statement-1 is False, Statement-2 is true

Answer

To solve the problem, we need to verify both statements and derive the necessary relationships step by step.

### Step 1: Understanding the Statements
- **Statement 1**: If \( \theta \) is acute and \( 1 + \cos \theta = k \), then \( \sin \left(\frac{\theta}{2}\right) = \sqrt{\frac{2 - k}{2}} \).
- **Statement 2**: \( 2 \sin^2 \left(\frac{x}{2}\right) = 1 - \cos x \).

### Step 2: Proving Statement 2
We start with Statement 2:
\[ 2 \sin^2 \left(\frac{x}{2}\right) = 1 - \cos x \]

Using the double angle formula for cosine:
\[ \cos x = 1 - 2 \sin^2 \left(\frac{x}{2}\right) \]

Rearranging gives:
\[ 1 - \cos x = 2 \sin^2 \left(\frac{x}{2}\right) \]

This confirms that Statement 2 is true.

### Step 3: Proving Statement 1
Now, we need to prove Statement 1. We start with the given equation:
\[ 1 + \cos \theta = k \]

From this, we can express \( \cos \theta \):
\[ \cos \theta = k - 1 \]

Next, we use the half-angle identity for sine:
\[ \sin \left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \cos \theta}{2}} \]

Substituting \( \cos \theta \) into the half-angle formula:
\[ \sin \left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - (k - 1)}{2}} \]
\[ = \sqrt{\frac{1 - k + 1}{2}} \]
\[ = \sqrt{\frac{2 - k}{2}} \]

This shows that Statement 1 is also true.

### Conclusion
Both statements are verified to be true:
- Statement 1: \( \sin \left(\frac{\theta}{2}\right) = \sqrt{\frac{2 - k}{2}} \) is correct.
- Statement 2: \( 2 \sin^2 \left(\frac{x}{2}\right) = 1 - \cos x \) is correct.

### Final Answer
Both statements are true.

---