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

0

The correct answer is Option (1) → 0

$Δ=\begin{vmatrix} -1 & \cos C & \cos B \\\cos C & -1 & \cos A\\\cos B & \cos A& -1\end{vmatrix}$

$A+B+C=π$

on expanding

$Δ=-1+2\cos A\cos B\cos C+\cos^2A+\cos^2B+\cos^2C$

so $2(\cos^2A+\cos^2B+\cos^2C)$

$=3+\cos^2A+\cos^2B+\cos^2C$

$=3+2\cos(A+B)\cos(A-B)+\cos 2C$

$=2+2\cos(π-C)\cos(A-B)+2\cos^2C$

$=2-2\cos C(\cos(A-B)-\cos C)$

$=2-2\cos C(\cos(A-B)+\cos(π-C))$

$2-2\cos C(\cos(A-B)+\cos(A+B))$

$=2-4\cos A\cos B\cos C$

so $Δ=-1+2\cos A\cos B\cos C+1-\cos A\cos B\cos C$

$=0$