Question

If \(A+B+C=\pi\), then \[ \begin{vmatrix} \sin(A+B+C) & \sin B & \cos C -\sin B & 0 & \tan A \cos(A+B) & -\tan A & 0 \end{vmatrix} \] is equal to

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

Hint:

Use identities like \(A+B+C=\pi\) → convert trig terms and look for cancellation.

Answer 1

The Correct Option is C

Concept: \(A+B+C=\pi \Rightarrow \sin(A+B+C)=\sin\pi=0\), and \(\cos(A+B)=\cos(\pi-C)=-\cos C\).

Step 1: Substitute values. Determinant becomes: \[ \begin{vmatrix} 0 & \sin B & \cos C -\sin B & 0 & \tan A -\cos C & -\tan A & 0 \end{vmatrix} \]

Step 2: Expand determinant. \[ = 0\cdot\begin{vmatrix}0 & \tan A -\tan A & 0\end{vmatrix} - \sin B \cdot \begin{vmatrix}-\sin B & \tan A -\cos C & 0\end{vmatrix} + \cos C \cdot \begin{vmatrix}-\sin B & 0 -\cos C & -\tan A\end{vmatrix} \]

Step 3: Simplify. \[ = -\sin B [(-\sin B)(0) - (\tan A)(-\cos C)] + \cos C [(-\sin B)(-\tan A) - 0] \] \[ = -\sin B (\tan A \cos C) + \cos C (\sin B \tan A) = 0 \]