Let `A=[[1,sintheta,1],[-sintheta,1,sintheta],[-1,-sintheta,1]]`, where `0lt=thetalt=2pi`. Then
(A) `D e t(A) = 0`
(B) `D e t(A) in (2,oo)`
(C) `D e t(A) in (2, 4)`
(D) `D e t (A) in [2, 4]`

Explanation

To find the determinant of the matrix \( A = \begin{bmatrix} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{bmatrix} \), we will follow these steps:

### Step 1: Write down the determinant formula
The determinant of a 3x3 matrix \( A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \) is calculated using the formula:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For our matrix, we have:
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