Question

If$A,B,C$ are acute positive angles such that $A+B+C=\pi$ and $cotAcotBcotC=K$ then

• A. $K\le \frac{1}{3\surd 3}$
• B. $K\ge \frac{1}{3\surd 3}$
• C. $K<\frac{1}{9}$
• D. $k>\frac{1}{3}$

Answer 1

The correct option is A

$K\le \frac{1}{3\surd 3}$

Explanation for correct option:

Step 1. Given that, $A,B,C$ are acute positive angles such that $A+B+C=\pi$

As we know,

$\begin{array}{rcl}\tan (\mathrm{A}+\mathrm{B}+\mathrm{C})& =& \frac{(\mathrm{tanA}+\mathrm{tanB}+\mathrm{tanC}–\mathrm{tanAtanBtan}\mathrm{C})}{(1–\mathrm{tanAtanB}–\mathrm{tanBtanC}–\mathrm{tanCtanA})}\end{array}\begin{array}{}\\ \\ \\ \end{array}$

$\Rightarrow$ $\tan \pi =\frac{(\tan A+\tan B+\tan C–\tan A\tan B\tan C)}{(1-\tan A\tan B–\tan B\tan C–\tan C\tan A)}$

$\Rightarrow$ $0=\frac{(\tan A+\tan B+\tan C–\tan A\tan B\tan C)}{(1–\tan A\tan B–\tan B\tan C–\tan C\tan A)}$

$\Rightarrow$ $0=\tan A+\tan B+\tan C–\tan A\tan B\tan C$

$\Rightarrow$$\tan A\tan B\tan C=\tan A+\tan B+\tan C\dots .(i)$

Step 2. finding value of K

As we know that Arithmetic mean $\ge$ Geometric mean

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

$\begin{array}{rcl}cotAcotBcotC& =& K\end{array}${given}

$\Rightarrow$$\begin{array}{rcl}\frac{1}{\tan A\tan B\tan C}& =& K\end{array}$

$\Rightarrow$$\begin{array}{rcl}\tan A\tan B\tan C& =& \frac{1}{K}.....\left(iii\right)\end{array}$

From equation (i), (ii), and (iii), we get

$\begin{array}{rcl}\frac{1}{3K}& \ge & K^{-\frac{1}{3}}\end{array}$

$\mathbf{\therefore }\mathit{K}\mathbf{\le }\frac{\mathbf{1}}{\mathbf{3}\sqrt{\mathbf{3}}}$

Hence, the correct option is $\left(A\right)$