Question

If in triangle ABC,$\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=0$ then $cotA+cotB+cotC=$

• A. $3\sqrt{3}$
• B. $\sqrt{3}$
• C. $\frac{1}{\sqrt{3}}$
• D. $\frac{1}{3\sqrt{3}}$

Answer 1

The correct option is B $\sqrt{3}$

$\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|$

$R_1\to R_1+R_2+R_3$

$a+b+c\left|\begin{array}{ccc}1& 1& 1\\ b& c& a\\ c& a& b\end{array}\right|$

$1(bc-a^2)-(b^2-ac)+ab-c^2=0$

$2\frac{1}{2}(a^2+b^2+c^2-ab-bc-ac)=0$

$(a-b{)}^2+(b-c{)}^2-(c-a{)}^2=0$

$\Rightarrow a-b=0$

$b-c=0$

$c-a=0$

$\Rightarrow a=b=c.$

$\therefore cotA=cotB=cotC=cot\left(60\right)$

according to the qs we have $3cotA$

$=\frac{3}{\sqrt{3}}$

$=\sqrt{3}$