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

• A. $\sin A$
• B. $\sin A \cos B$
• C. 0
• D. None of these

Hint:

The determinant of a skew-symmetric matrix of odd order is always zero. Always look for this structure after substituting known angle relations.

Answer 1

The Correct Option is C

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 $0$.