# If $A,B,C$ are angles of a triangle and $\begin{vmatrix} 1&1&1\\1+\sin A&1+\sin B&1+\sin C\\\sin A+\sin ^2A&\sin B+\sin^2B& \sin C+\sin^2C\end{vmatrix}=0$

$\begin{array}{1 1}(A)\;\text{Isosceles}&(B)\;\text{Equilateral}\\(C)\;\text{Right angled triangle}&(D)\;\text{None of these}\end{array}$

Applying $C_2-C_1$ and $C_3-C_1$ on the given determinant we get
$\begin{vmatrix} 1& 0& 0\\1+\sin A&\sin V-\sin A&\sin C-\sin A\\\sin A+\sin ^2A&\sin B-\sin A+\sin ^2B-\sin ^2A&\sin C-\sin A+\sin ^2C-\sin ^2A\end{vmatrix}=0$
$\Rightarrow (\sin B-\sin A)(\sin C-\sin A)\times \begin{vmatrix} 1&1\\1+\sin B+\sin A&1+\sin C+\sin A\end{vmatrix}=0$
$\Rightarrow (\sin B-\sin A)(\sin C-\sin A)(\sin C-\sin B)=0$
$\Rightarrow \sin A=\sin B$ or $(\sin C=\sin A)$ or $\sin C=\sin B$
$\Rightarrow A=B$ Or $B=C$ (or) $C=A$
$\Rightarrow$ triangle is isosceles.
Hence (A) is the correct answer.