Question

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

• A. $\tan A+\tan B+\tan C$
• B. $\cot A\cot B\cot C$
• C. ${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C$
• D. $0$

Answer 1

The correct option is D $0$

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