Question

If A, B, C are the angles of a triangle, then $\left|\begin{array}{ccc}{\sin }^{2}A& \cot A& 1\\ {\sin }^{2}B& \cot B& 1\\ {\sin }^{2}C& \cot C& 1\end{array}\right|=$ _____________.

Answer 1

Given: A, B, C are the angles of a triangle
Then, A + B + C = $\mathrm{\pi }$

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


$$\begin{gathered} ∆=\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| \\ \mathrm{Applying}{R}_2\to R_2-R_1\mathrm{and}{R}_3\to 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\\ \left(\sin B-\sin A\right)\left(\sin B+\sin A\right)& \frac{\cos B}{\sin B}-\frac{\cos A}{\sin A}& 0\\ \left(\sin C-\sin A\right)\left(\sin C+\sin A\right)& \frac{\cos C}{\sin C}-\frac{\cos A}{\sin A}& 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{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 \left(A-B\right)}{\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 \left(A-C\right)}{\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 \left(A-B\right)}{\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 \left(A-C\right)}{\sin A\sin C}& 0\end{array}\right| \\ =\left|\begin{array}{ccc}{\sin }^{2}A& \cot A& 1\\ \sin \left(A+B\right)\sin \left(B-A\right)& \frac{\sin \left(A-B\right)}{\sin A\sin B}& 0\\ \sin \left(A+C\right)\sin \left(C-A\right)& \frac{\sin \left(A-C\right)}{\sin A\sin C}& 0\end{array}\right| \\ =\left|\begin{array}{ccc}{\sin }^{2}A& \cot A& 1\\ \sin \left(\mathrm{\pi }-\mathrm{C}\right)\sin \left(B-A\right)& \frac{\sin \left(A-B\right)}{\sin A\sin B}& 0\\ \sin \left(\mathrm{\pi }-\mathrm{B}\right)\sin \left(C-A\right)& \frac{\sin \left(A-C\right)}{\sin A\sin C}& 0\end{array}\right|\left(\because A+B+C=\mathrm{\pi }\right) \\ =\left|\begin{array}{ccc}{\sin }^{2}A& \cot A& 1\\ \sin \left(\mathrm{C}\right)\sin \left(B-A\right)& \frac{\sin \left(A-B\right)}{\sin A\sin B}& 0\\ \sin \left(\mathrm{B}\right)\sin \left(C-A\right)& \frac{\sin \left(A-C\right)}{\sin A\sin C}& 0\end{array}\right| \\ \mathrm{Expanding}\mathrm{along}{C}_3 \\ =1\left(\sin \left(C\right)\sin \left(B-A\right)\times \frac{\sin \left(A-C\right)}{\sin A\sin C}-\sin \left(\mathrm{B}\right)\sin \left(C-A\right)\times \frac{\sin \left(A-B\right)}{\sin A\sin B}\right) \\ =\sin \left(B-A\right)\times \frac{\sin \left(A-C\right)}{\sin A}-\sin \left(C-A\right)\times \frac{\sin \left(A-B\right)}{\sin A} \\ =\left(-\sin \left(A-B\right)\right)\times \frac{\left(-\sin \left(C-A\right)\right)}{\sin A}-\sin \left(C-A\right)\times \frac{\sin \left(A-B\right)}{\sin A} \\ =\sin \left(A-B\right)\times \frac{\sin \left(C-A\right)}{\sin A}-\sin \left(C-A\right)\times \frac{\sin \left(A-B\right)}{\sin A} \\ =0 \end{gathered}$$

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