Question

Using properties of determinants, show that triangle $ABC$ is isosceles, if : $\left|\begin{array}{ccc}1& 1& 1\\ 1+\cos A& 1+\cos B& 1+\cos C\\ {\cos }^{2}A+\cos A& {\cos }^{2}B+\cos B& {\cos }^{2}B+\cos C\end{array}\right|=0$

Answer 1

$\left|\begin{array}{ccc}1& 1& 1\\ 1+\cos A& 1+\cos B& 1+\cos C\\ {\cos }^{2}A+\cos A& {\cos }^{2}B+\cos B& {\cos }^{2}B+\cos C\end{array}\right|=0$

$C_2\to C_2-C_3,C_3\to C_3-C_2$

$\left|\begin{array}{ccc}1& 0& 0\\ 1+\cos A& \cos B-\cos C& \cos C-\cos B\\ {\cos }^{2}A+\cos A& \cos B-\cos C& \cos C-\cos B\end{array}\right|=0$

$1\left[\right(\cos B-\cos C\left)\right(\cos C-\cos B)-(\cos B-\cos C\left)\right(\cos C-\cos B\left)\right]=0$

$\cos B-\cos C=0\Rightarrow \cos B=\cos C\Rightarrow B=C$

$\therefore B=C$

Therefore triangle is isoceles.