If `A,B` and `C` are angles of a triangle then the determinant
`|(-1,cosC,cosB),(cosC,-1,cosA),(cosB,cosA,-1)|` is equal to

Options

A. 0
B. `-1`
C. `1`
D. None of these

Explanation

To solve the determinant

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

we will follow these steps:

### Step 1: Write the determinant
We start with the determinant as given:

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

### Step 2: Apply column operations
We will apply the column operation \( C_1 \rightarrow C_1 + C_2 + C_3 \). This means we will add the second and third columns to the first column.

After performing this operation, the first column will be:

\[ C_1 = \begin{pmatrix} -1 + \cos C + \cos B \\ \cos C - 1 + \cos A \\ \cos B + \cos A - 1 \end{pmatrix} \]

So, the new determinant becomes:

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

### Step 3: Simplify the determinant
Now we will simplify the determinant. Notice that each entry in the first column can be expressed in terms of the angles of the triangle using the cosine rule.

Using the projection rule for triangles, we know:

- \( a = b \cos C + c \cos B \)
- \( b = c \cos A + a \cos C \)
- \( c = a \cos B + b \cos A \)

This means that the first column can be simplified further, leading to a cancellation of terms.

### Step 4: Evaluate the determinant
After simplification, we can evaluate the determinant. The determinant will ultimately simplify to 0 due to the properties of the angles in a triangle.

Thus, we find that:

\[ D = 0 \]

### Final Answer
The value of the determinant is:

\[ \boxed{0} \]