If $$f(\theta) = \begin{vmatrix} 1 & \cos\theta & 1 \\ -\sin\theta & 1 & -\cos\theta \\ -1 & \sin\theta & 1 \end{vmatrix}$$ and A and B are respectively the maximum and the minimum values of $$f(\theta)$$, then (A, B) is equal to:

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

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

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

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

$$f(\theta) = 2 + \sin2\theta + \cos2\theta$$

The range of the trigonometric part is $$[-\sqrt{2}, \sqrt{2}]$$

The required pair $$(A, B)$$ is $$(2 + \sqrt{2}, 2 - \sqrt{2})$$