If ABC is a triangle, then what is the value of the determinant \begin{vmatrix} \cos C & \sin B & 0 \\ \tan A & 0 & \sin B \\ 0 & \tan(B+C) & \cos C \end{vmatrix} ?

Explanation

Since A+B+C = \pi in a triangle, \tan(B+C) = \tan(\pi - A) = -\tan A. Expanding the determinant along the first row: \cos C(0 - \sin B(-\tan A)) - \sin B(\tan A \cos C - 0) + 0. This simplifies to \cos C \sin B \tan A - \sin B \tan A \cos C = 0.

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