If A + B + C = 0, then prove that

Given, $\left|\begin{array}{ccc}1 & \cos C & \cos B \\ \cos C & 1 & \cos A \\ \cos B & \cos A & 1\end{array}\right|$

On finding the determinant, we have

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

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

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

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

$\left[\because \cos ^{2} B-\sin ^{2} A=\cos (A+B) \cdot \cos (A-B)\right]$

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

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

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

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

$=\cos C(\cos C-\cos C) \quad($ As $\cos C=\cos (A+B))$

$=0$