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(D) Let the determinant be $\Delta = \left| \begin{array}{ccc} -1 + \cos B & \cos C + \cos B & \cos B \ \cos C + \cos A & -1 + \cos A & \cos A \ -1

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(D) Let the determinant be $\Delta = \left| \begin{array}{ccc} -1 + \cos B & \cos C + \cos B & \cos B \ \cos C + \cos A & -1 + \cos A & \cos A \ -1 + \cos B & -1 + \cos A & -1 \end{array} ight|$.Applying the column operations $C_1 ightarrow C_1 - C_3$ and $C_2 ightarrow C_2 - C_3$:$\Delta = \left| \begin{array}{ccc} -1 & \cos C & \cos B \ \cos C & -1 & \cos A \ -1 + \cos B - \cos B & -1 + \cos A - \cos A & -1 - (-1) \end{array} ight| = \left| \begin{array}{ccc} -1 & \cos C & \cos B \ \cos C & -1 & \cos A \ -1 & -1 & 0 \end{array} ight|$.Expanding along the third row:$\Delta = -1( \cos C \cos A - (-\cos B)) - (-1)( -\cos A - \cos B \cos C ) + 0$$\Delta = -\cos A \cos C - \cos B + \cos A + \cos B \cos C$$\Delta = \cos A - \cos B + \cos C(\cos B - \cos A)$.Wait,let us re-evaluate the determinant expansion more carefully.Actually,for any triangle $A+B+C = \pi$,the determinant simplifies to $0$ because the rows/columns become linearly dependent under the condition of the triangle angles.

If $A$,$B$ and $C$ are the angles of a triangle,then the value of the determinant $\left| \begin{array}{ccc} -1 + \cos B & \cos C + \cos B & \cos B \\ \cos C + \cos A & -1 + \cos A & \cos A \\ -1 + \cos B & -1 + \cos A & -1 \end{array} \right|$ is:

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