If A, B and C denote the angles of a triangle, then

$Δ=\begin{vmatrix}-1 & cos C & cos B\\cos C & -1 & cos A\\cos B & cos A & -2\end{vmatrix}$

is independent of

B

The correct answer is option (2) : B

Multiplying $C_1$ by a and then applying

$C_1→C_1+bC_2+cC_3,$ we get

$Δ=\begin{vmatrix}-a+bcos C + c cos B & cos C & cos B\\acos C -b - c cos A & -1 & cos A\\a cos B+b cos A - 2c & cos A & -2\end{vmatrix}$

$⇒Δ=\frac{1}{2}\begin{vmatrix}0 & cos C & cos B\\0 & -1 & cos A\\-c & cos A & -2\end{vmatrix}$

$⇒Δ=-\frac{c}{a}(cos \, C \, cos\, A + cos \, B )$

$⇒Δ=-\frac{c}{a}sin\, C\, sin \, A, $ which is independent of B.