Since \(A + B + C = \pi\): \(\sin(A+B+C) = \sin\pi = 0\) and \(\cos(A+B) = \cos(\pi - C) = -\cos C\).

The determinant becomes \(\begin{vmatrix} 0 & \sin B & \cos C \\ -\sin B & 0 & \tan A \\ -\cos C & -\tan A & 0 \end{vmatrix}\)

This is a skew-symmetric matrix of odd order (3×3). The determinant of any such matrix is 0.

(Direct expansion also gives \(-\sin B\tan A\cos C + \cos C\sin B\tan A = 0\).)

Answer: (A) 0