Question:
If $A + B + C = \pi$, then $\begin{vmatrix} \sin(A+B+C) & \sin B & \cos C \\ \sin B & 0 & \tan A \\ \cos(A+B) & \tan A & 0 \end{vmatrix}$ equals
If $A + B + C = \pi$, then $\begin{vmatrix} \sin(A+B+C) & \sin B & \cos C \\ \sin B & 0 & \tan A \\ \cos(A+B) & \tan A & 0 \end{vmatrix}$ equals
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The determinant of a skew-symmetric matrix of odd order is always zero. Always look for this structure after substituting known angle relations.
Updated On: May 2, 2026
- $\sin A$
- $\sin A \cos B$
- 0
- None of these
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
Substitute $A+B+C=\pi$ to simplify the entries, then check for a skew-symmetric structure.
Step 2: Detailed Explanation:
$\sin(A+B+C) = \sin\pi = 0$ and $\cos(A+B) = \cos(\pi-C) = -\cos C$.
The matrix becomes $\begin{pmatrix}0&\sin B&\cos C\\ \sin B&0&\tan A\\ -\cos C&\tan A&0\end{pmatrix}$, which is skew-symmetric of odd order (3).
The determinant of any skew-symmetric matrix of odd order is $0$.
Step 3: Final Answer:
The determinant equals $0$.
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