If `C = 2 cos theta`, then the value of the determinant `Delta = |(C,1,0),(1,C,1),(6,1,c)|`, is

Options

A. `(sin 4 theta)/(sin theta)`
B. `(2 sin^(2) 2 theta)/(sin theta)`
C. `4 cos^(2) theta (2 cos theta -1)`
D. none of these

Explanation

To solve the determinant \( \Delta = \begin{vmatrix} C & 1 & 0 \\ 1 & C & 1 \\ 6 & 1 & C \end{vmatrix} \) where \( C = 2 \cos \theta \), we will follow these steps:

### Step 1: Write the determinant
We start with the determinant:
\[ \Delta = \begin{vmatrix} C & 1 & 0 \\ 1 & C & 1 \\ 6 & 1 & C \end{vmatrix} \]

### Step 2: Expand the determinant
We can expand the determinant using the first row:
\[ \Delta = C \begin{vmatrix} C & 1 \\ 1 & C \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 6 & C \end{vmatrix} + 0 \begin{vmatrix} 1 & C \\ 6 & 1 \end{vmatrix} \]

### Step 3: Calculate the 2x2 determinants
1. Calculate \( \begin{vmatrix} C & 1 \\ 1 & C \end{vmatrix} \):
\[ = C \cdot C - 1 \cdot 1 = C^2 - 1 \]

2. Calculate \( \begin{vmatrix} 1 & 1 \\ 6 & C \end{vmatrix} \):
\[ = 1 \cdot C - 1 \cdot 6 = C - 6 \]

### Step 4: Substitute back into the determinant
Now substituting these values back into the determinant:
\[ \Delta = C(C^2 - 1) - (C - 6) \]
\[ = C^3 - C - C + 6 \]
\[ = C^3 - 2C + 6 \]

### Step 5: Substitute \( C = 2 \cos \theta \)
Now we substitute \( C = 2 \cos \theta \):
\[ \Delta = (2 \cos \theta)^3 - 2(2 \cos \theta) + 6 \]
\[ = 8 \cos^3 \theta - 4 \cos \theta + 6 \]

### Final Result
Thus, the value of the determinant is:
\[ \Delta = 8 \cos^3 \theta - 4 \cos \theta + 6 \]