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

To find the value of the determinant \(\Delta = \begin{vmatrix} c & 1 & 0 \\ 1 & c & 1 \\ 6 & 1 & c \end{vmatrix}\) given that \(c = 2\cos\theta\), we need to evaluate this determinant step-by-step using properties of determinants.

Let's break down the steps and find the determinant:

1. First, write out the determinant: \[ \Delta = \begin{vmatrix} c & 1 & 0 \\ 1 & c & 1 \\ 6 & 1 & c \end{vmatrix} \]
2. Use the method of cofactor expansion along the third column to find the determinant: \[ \Delta = c \cdot \begin{vmatrix} c & 1 \\ 1 & c \end{vmatrix} - 1 \cdot \begin{vmatrix} c & 1 \\ 6 & 1 \end{vmatrix} \]
3. Calculate the first minor determinant: \[ \begin{vmatrix} c & 1 \\ 1 & c \end{vmatrix} = c \cdot c - 1 \cdot 1 = c^2 - 1 \]
4. Calculate the second minor determinant: \[ \begin{vmatrix} c & 1 \\ 6 & 1 \end{vmatrix} = c \cdot 1 - 1 \cdot 6 = c - 6 \]
5. Substitute the values obtained into the formula and simplify: \[ \Delta = c(c^2 - 1) - (c - 6) = c^3 - c - c + 6 = c^3 - 2c + 6 \]
6. Substitute \(c = 2\cos\theta\): \[ \Delta = (2\cos\theta)^3 - 2(2\cos\theta) + 6 \] \[ = 8\cos^3\theta - 4\cos\theta + 6 \]
7. Simplify further if possible using trigonometric identities. However, this expression does not directly simplify neatly into the given options. Evaluating it further would not match any of the given options as stated.

Upon computing these steps, none of the given options matches. Thus, the correct answer is in fact: None of these.