Question

If A, B and C are the angles of a triangle, then

$\left|\begin{array}{ccc}-1+cosB& cosC+cosB& cosB\\ cosC+cosA& -1+cosA& cosA\\ -1+cosB& -1+cosA& -1\end{array}\right|=$

Answer 1

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

$C_1\to C_1-C_3;C\to C_2-C_3$

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

$=\frac{1}{abc}\left|\begin{array}{ccc}-a& a\cos C& a\cos B\\ bcosC& -b& b\cos A\\ c\cos B& c\cos A& -c\end{array}\right|$

Applying $R_1\to R_1+R_2+R_3$

$=\frac{1}{abc}\left|\begin{array}{ccc}0& 0& 0\\ bcosC& -b& b\cos A\\ c\cos B& c\cos A& -c\end{array}\right|$

$=0$