Question

If $A+B+C=\pi$, then the value of $\left|\begin{array}{ccc}\sin \left(A+B+C\right)& \sin \left(A+C\right)& \cos C\\ -\sin B& 0& \tan A\\ \cos \left(A+B\right)& \tan \left(B+C\right)& 0\end{array}\right|$ is equal to

(a) 0
(b) 1
(c) 2 sin B tan A cos C
(d) none of these

Answer 1

(a) 0

$$\begin{gathered} A+B+C=\mathrm{\pi } \\ \Rightarrow A+C=\mathrm{\pi }-B\mathit{,}\mathit{}\mathit{}A+B=\mathrm{\pi }-C\mathrm{and}B+C=\mathrm{\pi }-A \\ \mathrm{Thus}\mathrm{the}\mathrm{determinant}\mathrm{becomes} \\ \left|\sin \mathrm{\pi }\sin \left(\mathrm{\pi }-B\right)\cos C \\ -\sin B0\tan A \\ \cos \left(\mathrm{\pi }-C\right)\tan \left(\mathrm{\pi }-A\right)0\right| \\ =\left|0\sin B\cos C \\ -\sin B0\tan A \\ -\cos C-\tan A0\right|\left[\sin \mathrm{\pi }=0,\sin \left(\mathrm{\pi }-B\right)=B,\cos \left(\mathrm{\pi }-C\right)=-\cos C,\tan \left(\mathrm{\pi }-A\right)=-\tan A\right] \\ \mathrm{It}\mathrm{is}\mathrm{a}\mathrm{skew}\mathrm{symmetric}\mathrm{matrix}\mathrm{of}\mathrm{the}\mathrm{odd}\mathrm{order}3.\mathrm{Thus},\mathrm{by}\mathrm{property}\mathrm{of}\mathrm{determinants},\mathrm{we}\mathrm{get} \\ \left|∆\right|=0 \\ \Rightarrow \left|0\sin B\cos C \\ -\sin B0\tan A \\ -\cos C-\tan A0\right|=0 \end{gathered}$$