Question

Show that the $△ABC$ is an equilateral 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

We have , $△=\left[\begin{array}{ccc}1& 1& 1\\ 1+\cos A& 1+\cos B& 1+\cos C\\ {\cos }^{2}A+\cos A& {\cos }^{2}B+\cos B& {\cos }^{2}C+\cos C\end{array}\right]=0$

$[\because C_1\to C_1-C_3$ and $C_2\to C_2-C_3]$

$\Rightarrow (cosA-cosC).(cosB-cosC)$

$\left|\begin{array}{ccc}0& 0& 1\\ 1& 1& 1+cosC\\ cosA+cosC+1& cosB+cosC+1& co{s}^{2}C+cosC\end{array}\right|=0$

[ taking $(cosA-cosC)$ common from $C_1$ and $(cosB-cosC)$ common from $C_2$]

$\Rightarrow (cosA-cosC).(cosB-cosC)\left[\right(cosB+cosC+1)-(cosA+cosC+1\left)\right]=0$

$\Rightarrow (cosA-cosC).(cosB-cosC)(cosB+cosC+1-cosA-cosC-1)=0$

$\Rightarrow (cosA-cosC).(cosB-cosC)(cosB-cosA)=0$

i.e. $cosA=cosC$ or $cosB=cosC$ or $cosB=cosA$

$\Rightarrow A=C$ or $B=C$ or $B=A$

hence $ABC$ is an isoceles triangle.