Statement:1: The value of k for which `(costheta- sintheta)^(2) + k sintheta costheta-1=0` is an identity is `-2`.
and Statement -2: An identify in `theta` is satisfiied by all real values of `theta`.

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 determine the value of \( k \) for which the equation

\[ (\cos \theta - \sin \theta)^2 + k \sin \theta \cos \theta - 1 = 0 \]

is an identity in \( \theta \). An identity in \( \theta \) means that the equation holds true for all values of \( \theta \).

### Step 1: Expand the left-hand side
We start by expanding the square term:

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

Using the Pythagorean identity \( \cos^2 \theta + \sin^2 \theta = 1 \), we can rewrite this as:

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

### Step 2: Substitute back into the equation
Now, we substitute this back into the original equation:

\[ 1 - 2 \sin \theta \cos \theta + k \sin \theta \cos \theta - 1 = 0 \]

This simplifies to:

\[ -2 \sin \theta \cos \theta + k \sin \theta \cos \theta = 0 \]

### Step 3: Factor out the common term
Factoring out \( \sin \theta \cos \theta \):

\[ \sin \theta \cos \theta (k - 2) = 0 \]

### Step 4: Analyze the equation
For this equation to hold for all \( \theta \), the factor \( k - 2 \) must equal zero because \( \sin \theta \cos \theta \) is not zero for all \( \theta \). Therefore, we set:

\[ k - 2 = 0 \implies k = 2 \]

### Conclusion
Thus, the value of \( k \) for which the equation is an identity is \( k = 2 \), not \( -2 \) as stated in Statement 1. Therefore, Statement 1 is false, and Statement 2, which states that an identity in \( \theta \) is satisfied by all real values of \( \theta \), is true.