If $A+B+C=\pi$, then the value of $\left|\begin{array}{ccc}\sin (A+B+C) & \sin (A+C) & \cos C \\ -\sin B & 0 & \tan A \\ \cos (A+B) & \tan (B+C) & 0\end{array}\right|$ is equal to
(a) 0
(b) 1
(c) $2 \sin B \tan A \cos C$
(d) none of these
(a) 0
$A+B+C=\pi$
$\Rightarrow A+C=\pi-B, A+B=\pi-C$ and $B+C=\pi-A$
Thus the determinant becomes
$\mid \begin{array}{lll}\sin \pi & \sin (\pi-B) & \cos C\end{array}$
$\begin{array}{lll}-\sin B & 0 & \tan A\end{array}$
$\cos (\pi-C) \quad \tan (\pi-A) \quad 0$
$=\mid \begin{array}{lll} & 0 & \sin B & \cos C\end{array}$
$-\sin B \quad 0 \quad \tan A$
$-\cos C-\tan A \quad 0$ $[\sin \pi=0, \sin (\pi-B)=B, \cos (\pi-C)=-\cos C, \tan (\pi-A)=-\tan A]$
It is a skew symmetric matrix of the odd order 3 . Thus, by property of determinants, we get
$|\Delta|=0$
$\Rightarrow \mid \begin{array}{ccc}0 & \sin B & \cos C\end{array}$
$-\sin B \quad 0 \quad \tan A$
$\begin{array}{lll}-\cos C & -\tan A & 0\end{array}=0$