Question

If $\cot A+\cot B+\cot C=\sqrt{3}$, then the triangle $ABC$ is

• A. $equilateral$
• B. $rightangled$
• C. $isosceles$
• D. $noneofthese$

Answer 1

The correct option is A $equilateral$

We have, $\cot A+\cot B+\cot C=\sqrt{3}$

Squaring,

${\cot }^{2}A+{\cot }^{2}B+{\cot }^{2}C+2\cot A\cot C+2\cot B\cot C+2\cot C\cot A=3$ ...(1)

Now in $△ABC.$ $A+B+C=\pi \Rightarrow A+B=\pi -C$

$\Rightarrow \cot (A+B)=\cot (\pi -C)$

$\Rightarrow \frac{\cot A.\cot B-1}{\cot A+\cot B}=-\cot C$

$\Rightarrow \cot A\cot B+\cot B\cot C+\cot C\cot A=1$ ...(2)

From (1) and (2) , we have,

${\cot }^{2}A+{\cot }^{2}B+{\cot }^{2}C+2\cot A\cot B+2\cot B\cot C+2\cot C\cot A$

$=3[\cot A\cot B+\cot B\cot C+\cot C\cot A]$

$\Rightarrow {\cot }^{2}A+{\cot }^{2}B+{\cot }^{2}C-\cot A\cot B-\cot B\cot C-\cot C\cot A=0$

$\Rightarrow \frac{1}{2}[{(\cot A-\cot B)}^2+{(\cot B-\cot C)}^2+{(\cot C-\cot A)}^2]=0$

$[\because x^2+y^2+z^2-zx-xy-zy=\frac{1}{2}\{{(x-y)}^2+{(y-z)}^2+{(z-x)}^2\}]$

$\Rightarrow {(\cot A-\cot B)}^2+{(\cot B-\cot C)}^2+{(\cot C-\cot A)}^2=0$

$\Rightarrow \cot A=\cot B=\cot C$

[Since $R.H.S.$ is zero, each square must be zero]

$\Rightarrow A=B=C$ [for triangle]

$\Rightarrow$ Triangle is equilateral.