Question

Show that $△$ABC is an isosceles triangle, if the determinant

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

Answer 1

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

Applying $C_1\to C_1-C_3,C_2\to C_2-C_3$, we get,

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

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

Taking $(\cos A-\cos C)$ common from $C_1$ and $(\cos B-\cos C)$ common from $C_2$, we get

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

Expanding along $R_1$,

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

$(\cos A-\cos C)(\cos B-\cos C)(\cos B-\cos A)=0$

$\cos A=\cos Cor\cos B=\cos Cor\cos B=\cos A$

$A=CorB=CorB=A$

Hence, ABC is an isosceles triangle.