For triangle ABC,evaluate the determinant: le | vedclass

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(C) In $	riangle ABC$,we know that $A + B + C = \pi$,so $B + C = \pi - A$ and $A + C = \pi - B$. Substituting these into the determinant: $\sin(B + C

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$0$

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(C) In $	riangle ABC$,we know that $A + B + C = \pi$,so $B + C = \pi - A$ and $A + C = \pi - B$. Substituting these into the determinant: $\sin(B + C) = \sin(\pi - A) = \sin A$ $	an(A + C) = 	an(\pi - B) = -	an B$ The determinant becomes: $D = \left|\begin{array}{ccc}0 & \sin A & 	an B \ -\sin A & 0 & \cos C \ -	an B & -\cos C & 0\end{array}ight|$ This is a skew-symmetric determinant of order $3 	imes 3$. $A$ skew-symmetric matrix $M$ satisfies $M^T = -M$. The determinant of a skew-symmetric matrix of odd order $n$ is given by $\det(M) = \det(M^T) = \det(-M) = (-1)^n \det(M)$. Since $n = 3$ (odd),$\det(M) = -\det(M)$,which implies $2 \det(M) = 0$,so $\det(M) = 0$. Thus,the value of the determinant is $0$.

For $\triangle ABC$,evaluate the determinant: $\left|\begin{array}{ccc}0 & \sin A & \tan B \\ -\sin ( B + C ) & 0 & \cos C \\ \tan ( A + C ) & -\cos C & 0\end{array}\right|=$ . . . . . . .

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