\(\det(A)=1+\tan^2\theta=\sec^2\theta\) and \(\det(B)=1+\tan^2\theta=\sec^2\theta\). Since \(\det(C)=\det(AB^{-1})=\dfrac{\det(A)}{\det(B)}=\dfrac{\sec^2\theta}{\sec^2\theta}=1\).
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