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 :
Answer & explanation
Correct answer: option 1
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 $Δ=-1+2\cos A\cos B\cos C+1-\cos A\cos B\cos C$
$=0$