Question

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

Options

• 0

• 1

• 2 sin B tan A cos C

• none of these

Answer 1

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