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If $A, B, C$ are the angles of a triangle and $\left| {\begin{array}{*{20}{c}}1&1&1\\{1 + \sin A}&{1 + \sin B}&{1 + \sin C}\\{\sin A + {{\sin }^2}A}&{\sin B + {{\sin }^2}B}&{\sin C + {{\sin }^2}C} \end{array}} \right|$ $= 0$, then the triangle is
a equilateral
an isosceles
a right angled triangle
any triangle
Solution
$\left|\begin{array}{ccc}1 & 1 & 1 \\ 1+\sin A & 1+\sin B & 1+\sin C \\ \sin A+\sin ^{2} A & \sin B+\sin ^{2} B & \sin C+\sin ^{2} C\end{array}\right|=0$
$C_{2} \rightarrow C_{2}-C_{1}$
$\begin{array}{l}\Rightarrow\left|\begin{array}{ccc}1 & 0 & 1 \\ 1+\sin A & \sin B-\sin A & 1+\sin C \\ \sin A+\sin ^2 A & \sin B+\sin ^2 B-\sin A-\sin ^2 A & \sin C+\sin ^2 C\end{array}\right|=0 \\ C_3 \rightarrow C_3-C_1\end{array}$
$\begin{array}{l}\Rightarrow(\sin B-\sin A)(\sin C-\sin A) \left\lvert\, \begin{array}{ccc}1 & 0 & 0 \\ 1+\sin A & 1 & 1 \\ \sin A+\sin ^2 A & (\sin B+\sin A)+1 & (\sin C+\sin .\end{array}\right. \\\end{array}$
$\Rightarrow(\sin B-\sin A)(\sin C-\sin A)[(\sin C+\sin A)+1-(\sin B+\sin A)-1]=0$
$\Rightarrow(\sin B-\sin A)(\sin C-\sin A)[\sin C+\sin A)-\sin B-\sin A]=0$
$\Rightarrow(\sin B-\sin A)(\sin C-\sin A)(\sin C-\sin B)=0$
$\Rightarrow \sin B-\sin A=0$ or $\sin C-\sin A=0$ or $\sin C-\sin B=0$
$\Rightarrow \sin B=\sin A$ or $\sin C=\sin A$ or $\sin C=\sin B$
$\Rightarrow \mathrm{B}=\mathrm{A}$ or $\mathrm{C}=\mathrm{A}$ or $\mathrm{C}=\mathrm{B}$
$\triangle \mathrm{ABC}$ is an isosceles triangle.
