If A, B, C are the angles of a triangle,

Solution:

Given: $A, B, C$ are the angles of a triangle

Then, $A+B+C=\pi$

Let $\Delta=\left|\begin{array}{lll}\sin ^{2} A & \cot A & 1 \\ \sin ^{2} B & \cot B & 1 \\ \sin ^{2} C & \cot C & 1\end{array}\right|$

$\Delta=\left|\begin{array}{lll}\sin ^{2} A & \cot A & 1 \\ \sin ^{2} B & \cot B & 1 \\ \sin ^{2} C & \cot C & 1\end{array}\right|$

Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1}$

$=\left|\begin{array}{ccc}\sin ^{2} A & \cot A & 1 \\ \sin ^{2} B-\sin ^{2} A & \cot B-\cot A & 1-1 \\ \sin ^{2} C-\sin ^{2} A & \cot C-\cot A & 1-1\end{array}\right|$

$=\left|\begin{array}{ccc}\sin ^{2} A & \cot A & 1 \\ (\sin B-\sin A)(\sin B+\sin A) & \frac{\cos B}{\sin B}-\frac{\cos A}{\sin A} & 0 \\ (\sin C-\sin A)(\sin C+\sin A) & \frac{\cos C}{\sin C}-\frac{\cos A}{\sin A} & 0\end{array}\right|$

$=\left|\begin{array}{cccc}\sin ^{2} A & \cot A & 1 \\ \left(2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{B-A}{2}\right)\right)\left(2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{B-A}{2}\right)\right) & \frac{\cos B \sin A-\cos A \sin B}{\sin A \sin B} & 0 \\ \left(2 \cos \left(\frac{A+C}{2}\right) \sin \left(\frac{C-A}{2}\right)\right)\left(2 \sin \left(\frac{A+C}{2}\right) \cos \left(\frac{C-A}{2}\right)\right) & \frac{\cos C \sin A-\cos A \sin C}{\sin A \sin C} & 0\end{array}\right|$

$=\left|\begin{array}{ccc}\sin ^{2} A & \cot A & 1 \\ \left(2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A+B}{2}\right)\right)\left(2 \sin \left(\frac{B-A}{2}\right) \cos \left(\frac{B-A}{2}\right)\right) & \frac{\sin (A-B)}{\sin A \sin B} & 0 \\ \left(2 \cos \left(\frac{A+C}{2}\right) \sin \left(\frac{A+C}{2}\right)\right)\left(2 \sin \left(\frac{C-A}{2}\right) \cos \left(\frac{C-A}{2}\right)\right) & \frac{\sin (A-C)}{\sin A \sin C} & 0\end{array}\right|$

$=\left|\begin{array}{ccc}\sin ^{2} A & \cot A & 1 \\ \left(\sin 2\left(\frac{A+B}{2}\right)\right)\left(\sin 2\left(\frac{B-A}{2}\right)\right) & \frac{\sin (A-B)}{\sin A \sin B} & 0 \\ \left(\sin 2\left(\frac{A+C}{2}\right)\right)\left(\sin 2\left(\frac{C-A}{2}\right)\right) & \frac{\sin (A-C)}{\sin A \sin C} & 0\end{array}\right|$

$=\left|\begin{array}{ccc}\sin ^{2} A & \cot A & 1 \\ \sin (A+B) \sin (B-A) & \frac{\sin (A-B)}{\sin A \sin B} & 0 \\ \sin (A+C) \sin (C-A) & \frac{\sin (A-C)}{\sin A \sin C} & 0\end{array}\right|$

$=\left|\begin{array}{ccc}\sin ^{2} A & \cot A & 1 \\ \sin (\pi-C) \sin (B-A) & \frac{\sin (A-B)}{\sin A \sin B} & 0 \\ \sin (\pi-B) \sin (C-A) & \frac{\sin (A-C)}{-A} & 0\end{array}\right|$   $(\because A+B+C=\pi)$

$=\left|\begin{array}{ccc}\sin ^{2} A & \cot A & 1 \\ \sin (\mathrm{C}) \sin (B-A) & \frac{\sin (A-B)}{\sin A \sin B} & 0 \\ \sin (\mathrm{B}) \sin (C-A) & \frac{\sin (A-C)}{\sin A \sin C} & 0\end{array}\right|$

Expanding along $C_{3}$

$=1\left(\sin (C) \sin (B-A) \times \frac{\sin (A-C)}{\sin A \sin C}-\sin (\mathrm{B}) \sin (C-A) \times \frac{\sin (A-B)}{\sin A \sin B}\right)$

$=\sin (B-A) \times \frac{\sin (A-C)}{\sin A}-\sin (C-A) \times \frac{\sin (A-B)}{\sin A}$

$=(-\sin (A-B)) \times \frac{(-\sin (C-A))}{\sin A}-\sin (C-A) \times \frac{\sin (A-B)}{\sin A}$

$=\sin (A-B) \times \frac{\sin (C-A)}{\sin A}-\sin (C-A) \times \frac{\sin (A-B)}{\sin A}$

$=0$

Hence, if $A, B, C$ are the angles of a triangle, then $\left|\begin{array}{lll}\sin ^{2} A & \cot A & 1 \\ \sin ^{2} B & \cot B & 1 \\ \sin ^{2} C & \cot C & 1\end{array}\right|=\underline{0}$.