Choose the correct answer from the following :
The value of `|{:(-costheta,sintheta),(-sintheta,-costheta):}|`is:

Options

A. `-1`
B. 1
C. 0
D. None of these

Answer

To find the value of the determinant

\[ \left| \begin{array}{cc} -\cos \theta & \sin \theta \\ -\sin \theta & -\cos \theta \end{array} \right| \]

we can use the formula for a 2x2 determinant:

\[ \text{det} = ad - bc \]

where \( a, b, c, d \) are the elements of the matrix:

\[ \left| \begin{array}{cc} a & b \\ c & d \end{array} \right| = \left| \begin{array}{cc} -\cos \theta & \sin \theta \\ -\sin \theta & -\cos \theta \end{array} \right| \]

Here, we identify:
- \( a = -\cos \theta \)
- \( b = \sin \theta \)
- \( c = -\sin \theta \)
- \( d = -\cos \theta \)

Now, substituting these values into the determinant formula:

\[ \text{det} = (-\cos \theta)(-\cos \theta) - (\sin \theta)(-\sin \theta) \]

Calculating each term:

1. The first term:
\[ (-\cos \theta)(-\cos \theta) = \cos^2 \theta \]

2. The second term:
\[ (\sin \theta)(-\sin \theta) = -\sin^2 \theta \]

So we have:

\[ \text{det} = \cos^2 \theta + \sin^2 \theta \]

Using the Pythagorean identity, we know:

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

Thus, the value of the determinant is:

\[ \text{det} = 1 \]

### Final Answer:
The value of the determinant is \( 1 \).