# show that the $\Delta$ABC is an isosceles triangle if the determinant$\Delta=\begin{bmatrix}1 & 1& 1\\1+\cos A & 1+\cos B & 1+\cos C\\\cos^2A+\cos A & \cos^2B+\cos B & \cos^2C+\cos C\end{bmatrix}=0$

Toolbox:
• A square matrix A can be expressed as the sum of two or more terms,then |A| can be expressed as the sum of the determinant of two or more matrices.
• If A$=\begin{vmatrix}a_{11} & a_{21} & a_{31}\\a_{12} & a_{22} & a_{32}\\a_{13} & a_{23} & a_{33}\end{vmatrix}$
• Then |A|=$a_{11}(a_{22}\times a_{33}-a_{32}\times a_{23})-a_{21}(a_{12}\times a_{33}-a_{32}\times a_{13})+a_{31}(a_{12}\times a_{23}-a_{22}\times a_{13})$
Let $\Delta=\begin{bmatrix}1 & 1& 1\\1+\cos A & 1+\cos B & 1+\cos C\\\cos^2A+\cos A & \cos^2B+\cos B & \cos^2C+\cos C\end{bmatrix}$
Apply $C_2\rightarrow C_2-C_1$ and $C_3\rightarrow C_3-C_1$
$\Delta=\begin{bmatrix}1 & 0& 0\\1+\cos A & cos B-\cos A & \cos C-cos A\\\cos^2A+\cos A & \cos^2B-cos^2A+\cos B-cos A & \cos^2C-cos^2A+\cos C-cos A\end{bmatrix}$
But we know $cos^2B-cos^2A=(cos B+Cos A)(cos B-cos A)$
Therefore $\Delta=\begin{bmatrix}1 & 0& 0\\1+\cos A & cos B-\cos A & \cos C-cos A\\\cos^2A+\cos A & (\cos B-cos A)(cos B+cos A)+(\cos B-cos A) & (\cos C-cos A)(cos C+cos A)+(cos C-cos A)\end{bmatrix}$
Taking (cos B-cos A) from $C_2$ and (cos C-cos A) from $C_3$ as common factors we get,
$\Delta=(cos B-cos A)(cos C-cos A)\begin{vmatrix}1 & 0 & 0\\1+cos A & 1 & 1\\cos^2A+cos A& cos B+cos A&cos C+cos A\end{vmatrix}$
Now expanding along $R_1$ we get,
$\Delta=(cos B-cos A)(cos C-cos A)[1(cos C+cos A)-(cos B+cos A)]$
$\Delta=(cos B-cos A)(cos C-cos A)[cos C+cos A-cos B-cos A]$
$\quad=(cos B-cos A)(cos C-cos A)(cos C-cos B)$
Since $\Delta=0.$ is given,we can equate this to 0.
$\Delta=(cos B-cos A)(cos C-cos A)(cos C-cos B)=0$
$\Rightarrow cos B=cos A$ or cos C=cos A or cos C=cos B.
$\Rightarrow$ B=A or A=C or C=B.
Hence $\Delta$ABC is isosceles.