# If $A,B,C$ are the angles of a $\Delta$le then the value of the determinant $\small\begin{vmatrix}-1+\cos B&\cos C+\cos B&\cos B\\\cos C+\cos A&-1+\cos A&\cos A\\-1+\cos B&-1+\cos B&-1\end{vmatrix}$ is

$(a)\;0\qquad(b)\;1\qquad(c)\;-1\qquad(d)\;2$

We have
$\begin{vmatrix}-1+\cos B&\cos C+\cos B&\cos B\\\cos C+\cos A&-1+\cos A&\cos A\\-1+\cos B&-1+\cos B&-1\end{vmatrix}$
Apply $C_1\Rightarrow C_1-C_3$ and $C_2\rightarrow C_2-C_3$
$\begin{vmatrix}-1&\cos C&\cos B\\cos C&-1&\cos A\\\cos B&\cos A&-1\end{vmatrix}$
Multiply and divide by a in column 1
Apply $C_1\rightarrow C_1+bC_2+cC_3$
$\large\frac{1}{a}$$\begin{vmatrix}-a&\cos C&\cos B\\a\cos C&-1&\cos A\\acos B&\cos A&-1\end{vmatrix} \Rightarrow \large\frac{1}{a}$$\begin{vmatrix}0&\cos C&\cos B\\0&-1&\cos A\\0&\cos A&-1\end{vmatrix}$
$\Rightarrow 0$
Hence (a) is the correct answer.
edited Mar 20, 2014