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

To solve the determinant problem with the given condition \( A + B + C = \pi \), let's first apply this condition directly within the determinant.

Given matrix: \[ \begin{vmatrix} \sin(A+B+C) & \sin B & \cos C \\ \sin B & 0 & \tan A \\ \cos(A+B) & \tan A & 0 \end{vmatrix} \]

Using the identity \( \sin(A + B + C) = \sin(\pi) = 0 \), substitute this into the matrix:

\[ \begin{vmatrix} 0 & \sin B & \cos C \\ \sin B & 0 & \tan A \\ \cos(A+B) & \tan A & 0 \end{vmatrix} \]

The determinant of a matrix where one entire row or column contains all zeros is zero. In this case, the first column now contains only one non-zero element as 0, making the calculation simplify quickly. Therefore, the determinant of the matrix is:

0

Hence, the correct answer is 0, showcasing that the entire system leads to no valid output other than zero when computed utilizing the fundamental determinant properties.