Question

If \(c = 2\cos\theta\), then the value of the determinant \(\Delta = \begin{vmatrix} c & 1 & 0 \\ 1 & c & 1 \\ 6 & 1 & c \end{vmatrix}\) is

• A. \(\frac{\sin 4\theta}{\sin\theta}\)
• B. \(\frac{2\sin^2 2\theta}{\sin\theta}\)
• C. \(4\cos^2\theta(2\cos\theta - 1)\)
• D. None of these

Hint:

Use determinant expansion formula: \(|A| = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Expand the determinant and substitute \(c = 2\cos\theta\).
Step 2: Detailed Explanation:
\(\Delta = c(c^2 - 1) - 1(1 \cdot c - 6) + 0 = c^3 - c - c + 6 = c^3 - 2c + 6\)
\(= 8\cos^3\theta - 4\cos\theta + 6\)
This simplifies to \(2(4\cos^3\theta - 2\cos\theta + 3)\)
Step 3: Final Answer:
None of the given options match.