If A, B, C are the angles of a triangle and l | vedclass

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(B) Given the determinant: $\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+\s

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an isosceles triangle

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(B) Given the determinant: $\left|\begin{array}{ccc}1 & 1 & 1 \ 1+\sin A & 1+\sin B & 1+\sin C \ \sin A+\sin ^{2} A & \sin B+\sin ^{2} B & \sin C+\sin ^{2} C\end{array}ight|=0$Applying column operations $C_2 ightarrow C_2 - C_1$ and $C_3 ightarrow C_3 - C_1$:$\left|\begin{array}{ccc}1 & 0 & 0 \ 1+\sin A & \sin B-\sin A & \sin C-\sin A \ \sin A+\sin^2 A & (\sin B-\sin A)(1+\sin B+\sin A) & (\sin C-\sin A)(1+\sin C+\sin A)\end{array}ight|=0$Taking $(\sin B - \sin A)$ common from $C_2$ and $(\sin C - \sin A)$ common from $C_3$:$(\sin B - \sin A)(\sin C - \sin A) \left|\begin{array}{ccc}1 & 0 & 0 \ 1+\sin A & 1 & 1 \ \sin A+\sin^2 A & 1+\sin B+\sin A & 1+\sin C+\sin A\end{array}ight|=0$Expanding along the first row:$(\sin B - \sin A)(\sin C - \sin A) [ (1+\sin C+\sin A) - (1+\sin B+\sin A) ] = 0$$(\sin B - \sin A)(\sin C - \sin A)(\sin C - \sin B) = 0$This implies $\sin B = \sin A$ or $\sin C = \sin A$ or $\sin C = \sin B$.Since $A, B, C$ are angles of a triangle,this implies $A=B$ or $C=A$ or $C=B$.Therefore,the triangle is an isosceles triangle.

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

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