A determinant on a triangle

Let a triangle have angles $A, B, C$. Prove that:

$$\begin{vmatrix}
1 &\sin A  & \cot \frac{A}{2} \\\\
 1& \sin B & \cot \frac{B}{2}\\\\
 1& \sin C & \cot \frac{C}{2}
\end{vmatrix}=0$$

Solution

Maybe it is worth reading  Mollweide's formula

\begin{align*}
\mathcal{D} &=\begin{vmatrix}
1 & \sin A &\cot \frac{A}{2} \\\\
0 &\sin B - \sin A  &\cot \frac{B}{2}- \cot \frac{A}{2} \\\\
 0& \sin C - \sin B & \cot \frac{C}{2} - \cot \frac{B}{2}
\end{vmatrix} \\\\
 &= \begin{vmatrix}
\sin B - \sin A & \cot \frac{B}{2}- \cot \frac{A}{2} \\\\
\sin C - \sin A &  \cot \frac{C}{2} - \cot \frac{A}{2}
\end{vmatrix}\\\\
 &= \begin{vmatrix}
2 \sin \frac{B-A}{2} \cos \frac{B+A}{2}  &\frac{\sin \frac{B-A}{2}}{\sin \frac{B}{2} \sin \frac{A}{2}} \\\\
 2 \sin \frac{C-A}{2} \cos \frac{C+A}{2} & \frac{\sin \frac{C-A}{2}}{\sin \frac{C}{2} \sin \frac{A}{2}}
\end{vmatrix}\\\\
 &= 2 \sin \frac{B-A}{2} \sin \frac{C-A}{2} \begin{vmatrix}
\sin \frac{C}{2} &\frac{1}{\sin \frac{B}{2} \sin \frac{A}{2}} \\\\
 \sin \frac{B}{2}& \frac{1}{\sin \frac{C}{2} \sin \frac{A}{2}}
\end{vmatrix}\\
 &= 2 \sin \frac{B-A}{2} \sin \frac{C-A}{2} \frac{\sin \frac{C}{2} \sin \frac{B}{2} - \sin \frac{B}{2} \sin \frac{C}{2}}{\sin \frac{A}{2} \sin \frac{B}{2}\sin \frac{C}{2}} \\
 &=0
\end{align*}

The exercise can also be found at mathematica.gr