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
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
Show Hint
Use identities like \(A+B+C=\pi\) → convert trig terms and look for cancellation.
Updated On: Apr 18, 2026
- \(\sin A\)
- \(\sin A \cos B\)
- 0
- None of these
Show Solution
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The Correct Option is C
Solution and Explanation
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. \[ = -\sin B [(-\sin B)(0) - (\tan A)(-\cos C)] + \cos C [(-\sin B)(-\tan A) - 0] \] \[ = -\sin B (\tan A \cos C) + \cos C (\sin B \tan A) = 0 \]
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. \[ = -\sin B [(-\sin B)(0) - (\tan A)(-\cos C)] + \cos C [(-\sin B)(-\tan A) - 0] \] \[ = -\sin B (\tan A \cos C) + \cos C (\sin B \tan A) = 0 \]
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