If $A, B$ and $C$ are angles of a triangle, then the determinant $\begin{vmatrix} -1 & \cos C & \cos B \\ \cos C & -1 & \cos A \\ \cos B & \cos A & -1 \end{vmatrix}$ is equal to

$0$

The correct answer is Option (1) → $0$ ##

We have, $\begin{vmatrix} -1 & \cos C & \cos B \\ \cos C & -1 & \cos A \\ \cos B & \cos A & -1 \end{vmatrix}$

On applying $C_1 \rightarrow aC_1 + bC_2 + cC_3$, we get

$\begin{vmatrix} -a + b \cos C + c \cos B & \cos C & \cos B \\ a \cos C - b + c \cos A & -1 & \cos A \\ a \cos B + b \cos A - c & \cos A & -1 \end{vmatrix}$

Also, by projection rule in a triangle, we know that

$a = b \cos C + c \cos B, \quad b = c \cos A + a \cos C \quad \text{and} \quad c = a \cos B + b \cos A$

Using above equation in column first, we get

$\begin{vmatrix} -a + a & \cos C & \cos B \\ b - b & -1 & \cos A \\ c - c & \cos A & -1 \end{vmatrix} = \begin{vmatrix} 0 & \cos C & \cos B \\ 0 & -1 & \cos A \\ 0 & \cos A & -1 \end{vmatrix} = 0$  [since, all elements of column $C_1$ are zero]