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

$B - A = 1$

The correct answer is Option (4) → $B - A = 1$

We have:

$\begin{vmatrix} 1 & \cos\theta & 0 \\ \sin\theta & 1 & \cos\theta \\ \cos\theta & 1 & -\sin\theta \end{vmatrix}$

Expanding along the first row:

$= 1\begin{vmatrix} 1 & \cos\theta \\ 1 & -\sin\theta \end{vmatrix} - \cos\theta \begin{vmatrix} \sin\theta & \cos\theta \\ \cos\theta & -\sin\theta \end{vmatrix} + 0(\cdots)$

$= [1(-\sin\theta) - (\cos\theta)(1)] - \cos\theta \big[ (\sin\theta)(-\sin\theta) - (\cos\theta)(\cos\theta) \big]$

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

$= (-\sin\theta - \cos\theta) - \cos\theta(-1)$

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

$= -\sin\theta$

So, comparing with $A\sin\theta + B\cos\theta + C\sin\theta\cos\theta$:

$A = -1,\quad B = 0,\quad C = 0$

Thus, $B - A = 1 $