Question

lf $A+B+C=\pi$, then$\left|\begin{array}{ccc}{\sin }^{2}A& \cot A& 1\\ {\sin }^{2}B& \cot B& 1\\ {\sin }^{2}C& \cot C& 1\end{array}\right|$

• A. 0
• B. 1
• C. -1
• D. 2

Answer 1

The correct option is A 0

$\left|\begin{array}{ccc}{\sin }^{2}A& \cot A& 1\\ {\sin }^{2}B& \cot B& 1\\ {\sin }^{2}C& \cot C& 1\end{array}\right|$
$R_1\to R_1-R_{2,}{R}_2\to R_2-R_3$
$=\left|\begin{array}{ccc}{\sin }^{2}A-{\sin }^{2}B& \cot A-\cot B& 0\\ {\sin }^{2}B-{\sin }^{2}C& \cot B-\cot C& 0\\ {\sin }^{2}C& \cot C& 1\end{array}\right|$
$=\left|\begin{array}{ccc}\sin (A+B)\sin (A-B)& \frac{\cos A\sin B-\sin A\cos B}{\sin A\sin B}& 0\\ \sin (B+C)\sin (B-C)& \frac{\cos B\sin C-\sin B\cos C}{\sin B\sin C}& 0\\ {\sin }^{2}C& \cot C& 1\end{array}\right|$
$=\left|\begin{array}{ccc}\sin C\sin (A-B)& -\frac{\sin (A-B)}{\sin A\sin B}& 0\\ \sin A\sin (B-C)& -\frac{\sin (B-C)}{\sin B\sin C}& 0\\ {\sin }^{2}C& \cot C& 1\end{array}\right|$
$=-\frac{\sin (B-C)\sin (A-B)}{\sin B}+\frac{\sin (B-C)\sin (A-B)}{\sin B}$
$=0$