If A, B and C denote the angles of a triangle, then
`Delta = |(-1,cos C,cos B),(cos C,-1,cos A),(cos B,cos A,-2)|` is independent of

Options

A. A
B. B
C. C
D. none of these

Explanation

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

\[ \Delta = \begin{vmatrix} -1 & \cos C & \cos B \\ \cos C & -1 & \cos A \\ \cos B & \cos A & -2 \end{vmatrix} \]

### Step 1: Write the determinant
We can start by writing the determinant explicitly:

\[ \Delta = -1 \begin{vmatrix} -1 & \cos A \\ \cos A & -2 \end{vmatrix} - \cos C \begin{vmatrix} \cos C & \cos A \\ \cos B & -2 \end{vmatrix} + \cos B \begin{vmatrix} \cos C & -1 \\ \cos B & \cos A \end{vmatrix} \]

### Step 2: Calculate the 2x2 determinants
Now, we will calculate the 2x2 determinants:

1. For the first determinant:
\[ \begin{vmatrix} -1 & \cos A \\ \cos A & -2 \end{vmatrix} = (-1)(-2) - (\cos A)(\cos A) = 2 - \cos^2 A \]

2. For the second determinant:
\[ \begin{vmatrix} \cos C & \cos A \\ \cos B & -2 \end{vmatrix} = (\cos C)(-2) - (\cos A)(\cos B) = -2 \cos C - \cos A \cos B \]

3. For the third determinant:
\[ \begin{vmatrix} \cos C & -1 \\ \cos B & \cos A \end{vmatrix} = (\cos C)(\cos A) - (-1)(\cos B) = \cos C \cos A + \cos B \]

### Step 3: Substitute back into the determinant
Substituting these back into the expression for \(\Delta\):

\[ \Delta = -1(2 - \cos^2 A) - \cos C(-2 \cos C - \cos A \cos B) + \cos B(\cos C \cos A + \cos B) \]

### Step 4: Simplify the expression
Now we simplify the expression:

\[ \Delta = -2 + \cos^2 A + 2 \cos^2 C + \cos C \cos A \cos B + \cos B(\cos C \cos A + \cos B) \]

### Step 5: Use triangle properties
Using the properties of triangles, we know:
- \(a = b \cos C + c \cos B\)
- \(b = c \cos A + a \cos C\)
- \(c = a \cos B + b \cos A\)

### Step 6: Analyze independence
From the final expression of \(\Delta\), we can see that it contains terms involving \(\cos A\), \(\cos C\), and \(\cos B\). However, if we analyze the structure, we find that \(\Delta\) can be expressed in terms of \(A\) and \(C\) but not \(B\).

### Conclusion
Thus, we conclude that \(\Delta\) is independent of angle \(B\).