Question

If $f\left(x\right)=\frac{\tan x}{\tan 3x}$ then range of $f\left(x\right)$

• A. $(\frac{1}{3},1)$
• B. $(-\infty ,1)$
• C. $(-\infty ,\frac{1}{3})\cup (3,\infty )$
• D. $(\frac{1}{3},\infty )$

Answer 1

The correct option is C $(-\infty ,\frac{1}{3})\cup (3,\infty )$

$f\left(x\right)=\frac{\tan x}{\tan 3x}$

$=\frac{\tan x(1-3{\tan }^{2}x)}{3\tan x-{\tan }^{3}x}$
$y=\frac{1-3{\tan }^{2}x}{3-{\tan }^{2}x}$
$y=\frac{{\cos }^{2}x-3{\sin }^{2}x}{3{\cos }^{2}x-{\sin }^{2}x}$
$y=\frac{1-4{\sin }^{2}x}{3-4{\sin }^{2}x}$

Hence

$\frac{1+y}{1-y}=\frac{4-8{\sin }^{2}x}{2}$
$\frac{1+y}{2(1-y)}=1-2{\sin }^{2}x=\cos 2x$

Hence

$|\frac{1+y}{2(1-y)}|\le 1$

Hence

$y\le \frac{1}{3}$ and $y\ge 3$

$y\in (-\infty ,\frac{1}{3}]\cup [3,\infty )$