Question

For an $\Delta ABC$, the value of determinant $\left|\begin{array}{ccc}si{n}^{2}A& cotA& 1\\ si{n}^{2}B& cotB& 1\\ si{n}^{2}C& cotC& 1\end{array}\right|$ is equal to

• A. 0
• B. 1
• C. sin A sin B sin C
• D. sin A+ sin B+ sin C

Answer 1

Answer: (A)

0

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

det along $C_3$

$1({\sin }^{2}B\cot C-{\sin }^{2}C\cot B)-{\sin }^{2}A\cot C+{\sin }^{2}C\cot A+{\sin }^{2}A\cot B-\cot A{\sin }^{2}B$

A triangle meant it can even be an equilateral triangle

$\Rightarrow A=B=C=60°$

$\left|\begin{array}{ccc}\frac{3}{4}& \frac{1}{\sqrt{3}}& 1\\ \frac{3}{4}& \frac{1}{\sqrt{3}}& 1\\ \frac{3}{4}& \frac{1}{\sqrt{3}}& 1\end{array}\right|$

$\frac{\sqrt{3}}{4}\left|\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right|=0$

Option A