To find the maximum and minimum values of $f(\theta)$, we first need to calculate the determinant:

$f(\theta)=\left|\begin{array}{ccc}1 & \cos \theta & -1 \\ -\sin \theta & 1 & -\cos \theta \\ -1 & \sin \theta & 1\end{array}\right|$

We can expand the determinant along the first row:

$f(\theta) = 1 \cdot \left|\begin{array}{cc}1 & -\cos \theta \\ \sin \theta & 1\end{array}\right| - \cos \theta \cdot \left|\begin{array}{cc}-\sin \theta & -\cos \theta \\ -1 & 1\end{array}\right| + (-1) \cdot \left|\begin{array}{cc}-\sin \theta & 1 \\ -1 & \sin \theta\end{array}\right|$

Now, calculate each 2x2 determinant:

$\left|\begin{array}{cc}1 & -\cos \theta \\ \sin \theta & 1\end{array}\right| = (1)(1) - (-\cos \theta)(\sin \theta) = 1 + \sin \theta \cos \theta$

$\left|\begin{array}{cc}-\sin \theta & -\cos \theta \\ -1 & 1\end{array}\right| = (-\sin \theta)(1) - (-\cos \theta)(-1) = -\sin \theta - \cos \theta$

$\left|\begin{array}{cc}-\sin \theta & 1 \\ -1 & \sin \theta\end{array}\right| = (-\sin \theta)(\sin \theta) - (1)(-1) = -\sin^2 \theta + 1 = \cos^2 \theta$

Substitute these back into the expression for $f(\theta)$:

$f(\theta) = 1(1 + \sin \theta \cos \theta) - \cos \theta (-\sin \theta - \cos \theta) - 1(\cos^2 \theta)$

$f(\theta) = 1 + \sin \theta \cos \theta + \sin \theta \cos \theta + \cos^2 \theta - \cos^2 \theta$

$f(\theta) = 1 + 2 \sin \theta \cos \theta$

We know the trigonometric identity $2 \sin \theta \cos \theta = \sin(2\theta)$.

So, $f(\theta) = 1 + \sin(2\theta)$.

Now we need to find the maximum and minimum values of $f(\theta)$.

The range of the sine function $\sin(x)$ is $[-1, 1]$. Therefore, the range of $\sin(2\theta)$ is also $[-1, 1]$.

The minimum value of $f(\theta)$ occurs when $\sin(2\theta)$ is at its minimum, which is $-1$.

Minimum value $B = 1 + (-1) = 0$.

The maximum value of $f(\theta)$ occurs when $\sin(2\theta)$ is at its maximum, which is $1$.

Maximum value $A = 1 + (1) = 2$.

So, $A=2$ and $B=0$. The pair $(A,B)$ is $(2,0)$.

The calculated pair $(2,0)$ matches Option B.

The final answer is $\boxed{(2,0)}$

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