Let f(theta) = left| begin{array}{ccc} 1 & co | vedclass

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(B) To find $f(	heta)$, we expand the determinant along the first row: $f(	heta) = 1(1 - (-\sin 	heta \cdot -\cos 	heta)) - \cos 	heta(-\sin 	he

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$(2, 0)$

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(B) To find $f(	heta)$, we expand the determinant along the first row: $f(	heta) = 1(1 - (-\sin 	heta \cdot -\cos 	heta)) - \cos 	heta(-\sin 	heta - 1) - 1(-\sin^2 	heta + 1)$ $f(	heta) = 1(1 - \sin 	heta \cos 	heta) + \sin 	heta \cos 	heta + \cos 	heta + \sin^2 	heta - 1$ $f(	heta) = 1 - \sin 	heta \cos 	heta + \sin 	heta \cos 	heta + \cos 	heta + \sin^2 	heta - 1$ $f(	heta) = \sin^2 	heta + \cos 	heta$ Wait, let us re-evaluate the expansion: $f(	heta) = 1(1 + \sin 	heta \cos 	heta) - \cos 	heta(-\sin 	heta - \cos 	heta) - 1(-\sin^2 	heta + 1)$ $f(	heta) = 1 + \sin 	heta \cos 	heta + \sin 	heta \cos 	heta + \cos^2 	heta + \sin^2 	heta - 1$ $f(	heta) = 2 \sin 	heta \cos 	heta + (\sin^2 	heta + \cos^2 	heta) = \sin 2	heta + 1$ Since $-1 \le \sin 2	heta \le 1$, the range of $f(	heta)$ is $[1-1, 1+1] = [0, 2]$. Thus, the maximum value $A = 2$ and the minimum value $B = 0$. Therefore, $(A, B) = (2, 0)$.

Let $f(\theta) = \left| \begin{array}{ccc} 1 & \cos \theta & -1 \\ -\sin \theta & 1 & -\cos \theta \\ -1 & \sin \theta & 1 \end{array} \right|$. Suppose $A$ and $B$ are respectively the maximum and minimum values of $f(\theta)$. Then $(A, B)$ is equal to

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