Determinant doubt How do we prove, without actually expanding, that
$$\begin{vmatrix} 
\sin^2 {A}& \cot {A}& 1\\
\sin^2 {B}& \cot{B}& 1\\
\sin^2 {C}& \cot{C}& 1
\end{vmatrix}=0$$
where $A,B,C$ are angles of a triangle?
I tried applying cosine double angle formula but couldn't get anywhere. 

It is enough to show that
$$ \det\begin{pmatrix}\sin^3 A & \cos A & \sin A \\ \sin^3 B & \cos B & \sin B \\ \sin^3 C & \cos C & \sin C \end{pmatrix}=0 $$
or that
$$ \det\begin{pmatrix}a^3 & \frac{b^2+c^2-a^2}{bc} & a \\ b^3 & \frac{a^2+c^2-b^2}{ac} & b \\ c^3 & \frac{a^2+b^2-c^2}{ab} & c \end{pmatrix}=0 $$
or that
$$ \det\begin{pmatrix}a^3 b c & b^2+c^2-a^2 & abc \\ b^3 a c & a^2+c^2-b^2& abc \\ c^3ab & a^2+b^2-c^2 & abc \end{pmatrix}=0 $$
or that
$$ \det\begin{pmatrix}a^2 & b^2+c^2-a^2 & 1 \\ b^2 & a^2+c^2-b^2& 1 \\ c^2 & a^2+b^2-c^2 & 1 \end{pmatrix}=0 $$
or that
$$ \det\begin{pmatrix}a^2 & b^2+c^2 & 1 \\ b^2 & a^2+c^2& 1 \\ c^2 & a^2+b^2 & 1 \end{pmatrix}=0 $$
that is pretty simple.
By barycentric coordinates, this is equivalent to the collinearity of the centroid, symmedian point and symmedian point of the anticomplementary triangle (collinearity of $X(2),X(6)$ and $X(69)$ according to ETC). Pretty trivial since a triangle and its anticomplementary triangle have the same centroid.
