If $A + B + C = 0$, then prove that $\left| {\begin{array}{cccccccccccccccccccc}1&{\cos C}&{\cos B}\\{\cos C}&1&{\cos A}\\{\cos B}&{\cos A}&1\end{array}} \right| = 0$.

We have to prove, $\left| {\begin{array}{cccccccccccccccccccc}1&{\cos C}&{\cos B}\\{\cos C}&1&{\cos A}\\{\cos B}&{\cos A}&1\end{array}} \right| = 0$

$\therefore$ ${\rm{LHS}} = \left| {\begin{array}{cccccccccccccccccccc}1&{\cos C}&{\cos B}\\{\cos C}&1&{\cos A}\\{\cos B}&{\cos A}&1\end{array}} \right|$

$= 1\left( {1 - {{\cos }^2}A} \right) - \cos C(\cos C - \cos A \cdot \cos B) + \cos B(\cos C \cdot \cos A - \cos B)$

$= {\sin ^2}A - {\cos ^2}C + \cos A \cdot \cos B \cdot \cos C + \cos A \cdot \cos B \cdot \cos C - {\cos ^2}B$

$= {\sin ^2}A - {\cos ^2}B + 2\cos A \cdot \cos B \cdot \cos C - {\cos ^2}C$
$= - \cos (A + B) \cdot \cos (A - B) + 2\cos A \cdot \cos B \cdot \cos C - {\cos ^2}C$

$= - \cos ( - C) \cdot \cos (A - B) + \cos C(2\cos A \cdot \cos B - \cos C)$

$= - \cos C(\cos A \cdot \cos B + \sin A \cdot \sin B - 2\cos A \cdot \cos B + \cos C)$
$= \cos C(\cos A \cdot \cos B - \sin A \cdot \sin B - \cos C)$

$= \cos C[\cos (A + B) - \cos C]$
$= \cos C(\cos C - \cos C) = 0 = RHS$