<s>[INST] If $A$, $B$ and $C$ are the angles of a triangle then find the value of $\Delta$ I'll state the question from my book below:

If $A$, $B$ and $C$ are the angles of a triangle, then find the determinant value of
  $$\Delta = \begin{vmatrix}\sin^2A & \cot A & 1 \\ \sin^2B & \cot B & 1 \\ \sin^2C & \cot C & 1\end{vmatrix}.$$

Here's how I tried solving the problem:
$\Delta = \begin{vmatrix}\sin^2A & \cot A & 1 \\ \sin^2B & \cot B & 1 \\ \sin^2C & \cot C & 1\end{vmatrix}$
$R_2 \to R_2 - R_1$
$R_3 \to R_3 -R_1$
$= \begin{vmatrix}\sin^2A & \cot A & 1 \\ \sin^2B-\sin^2A & \cot B-\cot A & 0 \\ \sin^2C-\sin^2A & \cot C-\cot A & 0\end{vmatrix}$
Expanding the determinant along $C_3$
\begin{align}
&= (\sin^2B-\sin^2A)(\cot C-\cot A)-(\cot B-\cot A)(\sin^2C-\sin^2A)  \\
&= \sin(B+A) \sin(B-A) \left[\frac {\cos C}{\sin C} - \frac {\cos A}{\sin A}\right] - \left[\frac {\cos B}{\sin B} - \frac {\cos A}{\sin A}\right]\sin(C+A) \sin(C-A)  \\
&= \frac {\sin(B+A) \sin(B-A) \sin(A-C)} {\cos A \cos C} - \frac {\sin(A-B) \sin(C+A) \sin(C-A)} {\cos A \cos C}  \\
&= \frac {\sin(B-A) \sin (A-C)} {\cos A} \left[\frac {\sin(A+B)} {\cos C} - \frac {\sin(A+C)} {\cos B}\right]  \\
&= \frac {\sin(B-A) \sin (A-C)} {\cos A} \left[\frac {\sin C} {\cos C} - \frac {\sin B} {\cos B}\right]  \\
&= \frac {\sin(B-A) \sin (A-C) \sin (C-B)} {\cos A \cos B \cos C}
\end{align}
I tried solving further but the expression just got complicated. I don't even know if the work I've done above is helpful. My textbook gives the answer as $0$. I don't have any clue about getting the answer. Any help would be appreciated.
 [/INST] By Euler's theorem $O,G,H$ are collinear. By considering their trilinear coordinates it follows that
$$ \det\begin{pmatrix}\cos(A) & \cos(B) & \cos(C) \\ \frac{1}{\sin A}&\frac{1}{\sin B}&\frac{1}{\sin C}\\ \frac{1}{\cos A}&\frac{1}{\cos B}&\frac{1}{\cos C}\end{pmatrix}=0$$
and by multiplying the first column by $\cos(A)$, the second column by $\cos(B)$ and the third column by $\cos(C)$ we get that
$$ \det\begin{pmatrix}\cos^2(A) & \cos^2(B) & \cos^2(C) \\ \cot(A)&\cot(B)&\cot(C)\\ 1&1&1\end{pmatrix}=0$$
and by replacing the first row with the difference between the third row and the first row
$$ \det\begin{pmatrix}\sin^2(A) & \sin^2(B) & \sin^2(C) \\ \cot(A)&\cot(B)&\cot(C)\\ 1&1&1\end{pmatrix}=0$$
readily follows. One may also notice that if $\theta\in\{A,B,C\}$ then
$$ 8R^2\cdot \sin^2\theta + 4[ABC]\cdot\cot\theta -(a^2+b^2+c^2)\cdot 1 = 0 $$
hence the nullspace of the given matrix has dimension $\geq 1$.
</s>