If $\begin{vmatrix} 1 & \cos \theta & 0 \\ \sin \theta & 1 & \cos \theta \\ |\cos \theta & 1 & -\sin \theta| \end{vmatrix} = A\sin \theta + B\cos \theta + C\sin \theta\cos \theta$ then:

The determinant $\begin{vmatrix} 1 & \cos\theta & 0 \\ \sin\theta & 1 & \cos\theta \\ \cos\theta & 1 & -\sin\theta \end{vmatrix}$ equals $A\sin\theta + B\cos\theta + C\sin\theta\cos\theta$.

Expanding along Row 1 (since it contains a 0):

$\text{Det} = 1 \cdot \begin{vmatrix} 1 & \cos\theta \\ 1 & -\sin\theta \end{vmatrix} - \cos\theta \cdot \begin{vmatrix} \sin\theta & \cos\theta \\ \cos\theta & -\sin\theta \end{vmatrix} + 0 \cdot \begin{vmatrix} \sin\theta & 1 \\ \cos\theta & 1 \end{vmatrix}$

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The first 2×2 determinant:

$\begin{vmatrix} 1 & \cos\theta \\ 1 & -\sin\theta \end{vmatrix} = (1)(-\sin\theta) - (\cos\theta)(1)$

$= -\sin\theta - \cos\theta$

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The second 2×2 determinant:

$\begin{vmatrix} \sin\theta & \cos\theta \\ \cos\theta & -\sin\theta \end{vmatrix} = (\sin\theta)(-\sin\theta) - (\cos\theta)(\cos\theta)$

$= -\sin^2\theta - \cos^2\theta$

$= -(\sin^2\theta + \cos^2\theta)$

$= -1$

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Substituting back:

$\text{Det} = 1(-\sin\theta - \cos\theta) - \cos\theta(-1)$

$= -\sin\theta - \cos\theta + \cos\theta$

$= -\sin\theta$

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Comparing $-\sin\theta$ with $A\sin\theta + B\cos\theta + C\sin\theta\cos\theta$:

Coefficient of $\sin\theta$: $A = -1$

Coefficient of $\cos\theta$: $B = 0$

Coefficient of $\sin\theta\cos\theta$: $C = 0$

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Checking the options:

$A + B = -1 + 0 = -1 \neq 0$

$B + C = 0 + 0 = 0 \neq 2$

$C + A = 0 + (-1) = -1 \neq 1$

$B - A = 0 - (-1) = 1$ ✓

Therefore, $B - A = 1$.