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  4. If A + B + C = π, then the value | sin(A+B+C)& sin B& cos C - sin B&0& tan A cos(A+B)&- tan A&0|

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Q. If A + B + C = $\pi$, then the value $\begin{vmatrix}\sin\left(A+B+C\right)&\sin B&\cos C\\ -\sin B&0&\tan A\\ \cos\left(A+B\right)&-\tan A&0\end{vmatrix} $

Determinants

Solution:

The given determinant is skew-symmetric of odd order (namely 3),
( $\because$ cos (A+ B) = cos ($\pi$ - C) = - cos C and sin (A + B + C) = sin $\pi$ = 0)
therefore. if $\Delta$ is the value of determinant, then $\Delta = (-1)^3 \Delta \, \Rightarrow \, 2 \Delta = 0 $