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

$B$

Multiplying $C_1$ by a and then applying $C_1→C_1+b C_2 + c C_3$, we get

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

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

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

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