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

To solve the given problem, we need to evaluate the determinant of the matrix:

\(\sin(A+B+C)\)	\(\sin B\)	\(\cos C\)
-\(\sin B\)	0	\(\tan A\)
\(\cos(A+B)\)	-\(\tan A\)	0

We are given that \(A + B + C = \pi\). This fact allows us to simplify some trigonometric expressions:

• \(\sin(A+B+C) = \sin \pi = 0\)
• \(\cos(A+B+C) = \cos \pi = -1\)

Thus, the first element of the first row, \(\sin(A+B+C)\), becomes 0. The matrix reduces to:

0	\(\sin B\)	\(\cos C\)
-\(\sin B\)	0	\(\tan A\)
\(\cos(A+B)\)	-\(\tan A\)	0

Now, we can compute the determinant of this matrix using the standard determinant procedure for a \(3 \times 3\) matrix \( |M| \):

\( |M| = a(ei-fh) - b(di-fg) + c(dh-eg) \)

Substituting \( a = 0 \), \( b = \sin B \), \( c = \cos C \), etc., into the formula:

\(|M| = 0 \cdot (0 \cdot 0 - \tan A \cdot (-\tan A)) - \sin B \cdot (-\sin B \cdot 0 - \tan A \cdot \cos(A+B)) + \cos C \cdot (-\sin B \cdot (-\tan A) - 0 \cdot \cos(A+B)) \)

This simplifies to:

\(|M| = -\sin B \cdot (-\tan A \cdot \cos(A+B)) + \cos C \cdot (\sin B \cdot \tan A) \)

\(|M| = \sin B \tan A \cos(A+B) + \cos C \sin B \tan A \)

Since \(\tan A\) is common, factored out:

\(|M| = \tan A \sin B (\cos(A+B) + \cos C) \)

However, we know \(\cos(A+B) = \cos(\pi - C) = -\cos C\), so:

\(|M| = \tan A \sin B (\cos C - \cos C) = 0\)

Thus, the value of the determinant is zero.

Conclusion: The determinant of the given matrix is equal to 0.