<s>[INST] If $A+B+C=0$, then prove that the value of the determinant is $0$. I'll state the question from my textbook below:

If $A+B+C=0$, the prove that $\begin{vmatrix}1 & \cos C & \cos B \\ \cos C & 1 & \cos A \\ \cos B & \cos A & 1 \end{vmatrix} = 0$.

This is how I tried solving the problem:
$LHS = \begin{vmatrix}1 & \cos C & \cos B \\ \cos C & 1 & \cos A \\ \cos B & \cos A & 1 \end{vmatrix}$
$= 1(1- \cos^2 A) - \cos C (\cos C - \cos A \cos B) + \cos B (\cos A \cos C - \cos B)$
$= 1 + 2 \cos A \cos B \cos C - (\cos^2 A +\cos^2 B + \cos^2 C)$
I don't know how to proceed further. I tried using the fact that $A+B+C=0$ but it didn't lead to anything I could solve. I don't know where is it supposed to be used. 
Also, I read a solution to this problem somewhere in which the term $(\cos^2 A +\cos^2 B + \cos^2 C)$ was replaced by $1 + 2 \cos A \cos B \cos C$ as $A+B+C=0$. Are these two terms equal for the given condition? Also, is there a way to prove the statement without using this fact?
Any help would be appreciated.
 [/INST] If one of the angles, say $A=0$, then two rows become identical and the determinant becomes zero.
Else we multiply the rows in order by $\sin A, \sin B, \sin C$ and we get
\begin{vmatrix}\sin A & \sin A\cos C & \sin A\cos B \\ \sin B\cos C & \sin B & \sin B\cos A \\ \sin C\cos B & \sin C\cos A & \sin C \end{vmatrix}
Now with $R_1 \rightarrow R_1+R_2+R_3$ we see for instance that $$a_{11} = \sin A +\sin B \cos C+\sin C \cos B = \sin A+\sin (B+C) = \sin A - \sin A =0$$
i.e. the first row becomes null and hence the determinant equals zero.
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