If A,B,C are the angles of triangle ABC, then the minimum value of `|{:(-1,cos C , cos B),(cos C , -1, cos A ) , (cos B , cos A , -1):}|` is equal to :

Options

A. 0
B. `-1`
C. 1
D. `-2`

Answer

To find the minimum value of the determinant

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

where \(A, B, C\) are the angles of triangle \(ABC\), we will follow these steps:

### Step 1: Apply a Column Transformation

We will perform a column transformation by replacing the first column with \(A \cdot C_1 + B \cdot C_2 + C \cdot C_3\). This means we will multiply each column by the corresponding angle and sum them up.

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

### Step 2: Simplify the Determinant

After applying the column transformation, we will simplify the determinant. The first column will become:

\[ A \cdot (-1) + B \cdot \cos C + C \cdot \cos B, \]
\[ B \cdot \cos C + C \cdot \cos B - A, \]
\[ A \cdot \cos C + C \cdot \cos A - B, \]
\[ A \cdot \cos B + B \cdot \cos A - C. \]

### Step 3: Use the Projection Theorem

Using the projection theorem for triangle \(ABC\), we have:

\[ A = B \cos C + C \cos B, \]
\[ B = A \cos C + C \cos A, \]
\[ C = A \cos B + B \cos A. \]

### Step 4: Substitute Values

Substituting these values into the determinant, we will find that the first column becomes:

\[ 0, 0, 0. \]

### Step 5: Evaluate the Determinant

Since the first column consists entirely of zeros, the determinant evaluates to zero:

\[ D = 0. \]

### Conclusion

Thus, the minimum value of the determinant is

\[ \boxed{0}. \]

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