Question

If $A,B,C$ are acute angles such that $\tan A+\tan B+\tan C=\tan A\tan B\tan C$. Then $cotAcotBcotC$ is

• A. $\le \frac{1}{\sqrt{3}}$
• B. $\le \frac{1}{2\sqrt{3}}$
• C. $\le \frac{1}{3\sqrt{3}}$
• D. None of these

Answer 1

The correct option is C

$\le \frac{1}{3\sqrt{3}}$

Explanation for correct option.

Step1. Using relation between Arithmetic mean & Geometric mean

Given, $A,B,C$ are acute angles

$\because$Arithmetic mean of $3$ quantities $\ge$Geometric mean of $3$ quantities

⇒ $\begin{array}{rcl}\frac{[\tan A+\tan B+\tan C]}{3}& \ge & {\left(\tan A\tan B\tan C\right)}^{\frac{1}{3}}\end{array}$

Taking cube both sides,

⇒$\begin{array}{rcl}\frac{{[\tan A+\tan B+\tan C]}^3}{27}& \ge & \left(\tan A\tan B\tan C\right)\end{array}$

Step2. Finding value of $\mathit{c}\mathit{o}\mathit{t}\mathit{A}\mathit{c}\mathit{o}\mathit{t}\mathit{B}\mathit{c}\mathit{o}\mathit{t}\mathit{C}$

Also given,

$\begin{array}{rcl}\tan A+\tan B+\tan C& =& \tan AtanB\tan C\end{array}$

⇒$\begin{array}{rcl}\frac{{\left[\tan AtanB\tan C\right]}^3}{27}& \ge & \left(\tan A\tan B\tan C\right)\end{array}$

Divide both side by $\begin{array}{rcl}& & \tan AtanB\tan C\end{array}$

⇒$\begin{array}{rcl}{\left[\tan AtanB\tan C\right]}^2& \ge & 27\end{array}$

Taking square root both sides,

⇒$\begin{array}{rcl}\left[\tan AtanB\tan C\right]& \ge & 3\sqrt{3}\end{array}$

$\begin{array}{rcl}\therefore cotAcotBcotC& \le & \frac{1}{3\sqrt{3}}\end{array}$

Hence, correct option is $\left(C\right)$.