Question

Show that $\Delta ABC$ is an isosceles triangle, if the determinant
$\Delta =\left|\begin{array}{ccc}1& 1& 1\\ 1+cosA& 1+cosB& 1+cosC\\ co{s}^{2}A+cosA& co{s}^{2}B+cosB& co{s}^{2}C+cosC\end{array}\right|=0$

Answer 1

We have, $\Delta =\left|\begin{array}{ccc}0& 0& 1\\ cosA-cosC& cosB-cosC& 1+cosC\\ co{s}^{2}A+cosA-co{s}^{2}A-co{s}^{2}C-cosC& co{s}^{2}B+cosB-co{s}^{2}C-cosC& co{s}^{2}C+cosC\end{array}\right|=0$
$[\therefore C_1\to C_1-C_3\text{and}{C}_2\to C_2-C_3]$
$\Rightarrow (cosA-cosC).(cosB-cosC)\left|\begin{array}{ccc}0& 0& 1\\ 1& 1& 1+cosC\\ cosA+cosC+1& cosB+cosC+1& co{s}^{2}C+cosC\end{array}\right|=0$
[taking $(cosA-cosC)$ common from $C_1$ and $(cosB-cosC)$ common from $C_2$]
$\Rightarrow (cosA-cosC).(cosB-cosC)\left[\right(cosB+cosC+1)-(cosA+cosC+1\left)\right]=0\Rightarrow (cosA-cosC).(cosB-cosC)(cosB+cosC+1-cosA-cosC-1)=0\Rightarrow (cosA-cosC).(cosB-cosC)(cosB-cosA)=0i.e.,cosA=cosCorcosB=cosCorcosB=cosA$
$\Rightarrow A=C\text{or}B=C\text{or}B=A$
Hence, ABC is an isosceles triangle.