Question

Show that the value of $\frac{\tan x}{\tan 3x}$ wherever defined never lies between $\frac{1}{3}$ and $3$.

Answer 1

$k=\frac{\tan x}{\tan 3x}=\frac{\tan x}{\tan (x+2x)}$

$=\frac{\tan x}{\frac{\tan 2x+\tan x}{1-\tan 2x\left(\tan x\right)}}$

$\frac{\left[\tan x\right][1-\tan x(\frac{2\tan x}{1-\tan x}\left)\right]}{\frac{2\tan x}{1-{\tan }^{2}x}+\tan x}$

$=\frac{\left(\tan x\right)[1-{\tan }^{2}x-2{\tan }^{2}x]}{\left(\tan x\right)[2+1-{\tan }^{2}x]}$

$k=\frac{1-3{\tan }^{2}x}{3-{\tan }^{2}x}$ Put $\tan x=t$

$k=\frac{1-3t^2}{3-t^2}\Rightarrow 3k-k{t}^2=1-3t^2$

$=t^2(3-k)=1-3k$

$=t^2=\frac{(1-3k)}{(3-k)}$

$f^2>0$

$\Rightarrow \frac{(3k-1)}{(3-k)}>0$

$\Rightarrow \frac{(3k-1)(3-k)}{(3-k{)}^2}>0$

$\Rightarrow (3k-1)(3-k)>0$

$\Rightarrow k\in (-\infty ,\frac{1}{3})\cup (3,\infty )$

$\frac{\tan x}{\tan 3x}\in (-\infty ,\frac{1}{3})\cup (3,\infty )$