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Question:

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

Solution:

(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

$|\Delta|=0$

$\Rightarrow \mid \begin{array}{ccc}0 & \sin B & \cos C\end{array}$

$-\sin B \quad 0 \quad \tan A$

$\begin{array}{lll}-\cos C & -\tan A & 0\end{array}=0$