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  4. If x+y+z=π then the value of | sin (x+y+z) sin B cos z - sin B 0 tan A cos (x+y) - tan A 0| is

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Q. If $x+y+z=\pi$ then the value of $\begin{vmatrix}\sin (x+y+z) & \sin B & \cos z \\ -\sin B & 0 & \tan A \\ \cos (x+y) & -\tan A & 0\end{vmatrix}$ is

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Solution:

The given determinant $=\begin{vmatrix}\sin (\pi) & \sin B & \cos z \\ -\sin B & 0 & \tan A \\ \cos (\pi- z ) & -\tan A & 0\end{vmatrix}$
$=\begin{vmatrix}0 & \sin B & \cos z \\ -\sin B & 0 & \tan A \\ -\cos z & -\tan A & 0\end{vmatrix}=0$
$\because$ the corresponding matrix is skew-symmetric.