Question

If $A+B+C=\pi$, prove that $\left|\begin{array}{ccc}{\sin }^{2}A& \cot A& 1\\ {\sin }^{2}B& \cot B& 1\\ {\sin }^{2}C& \cot C& 1\end{array}\right|=0$.

Answer 1

Apply $R_1-R_2,R_2-R_3$ to make two zeros.
Use ${\sin }^{2}A-{\sin }^{2}B-\sin (A+B)\sin (A-B)=\sin C\sin (A-B)$
and $\cot A-\cot B=\frac{\sin (B-A)}{\sin A\sin B}$
$=\frac{-\sin C\sin (A-B)}{\sin A\sin B\sin C}$
$\Delta =\left|\begin{array}{cc}-\sin C\sin (A-B)& -\sin C\sin (A-B)\\ -\sin A\sin (B-C)& -\sin A\sin (B-C)\end{array}\right|÷\sin A\sin B\sin C$
Clearly $\Delta =0$ as the two rows are identical, i.e.,
$\Delta =\lambda \left|\begin{array}{cc}1& 1\\ 1& 1\end{array}\right|=0$.