Question

The value of $\frac{\tan x}{\tan 3x}$ wherever defined never lies between

• A. $\frac{1}{2}$ and 2
• B. $\frac{1}{3}$ and 3
• C. $\frac{1}{4}$ and 4
• D. $\frac{1}{3}$ and 2

Answer 1

Answer: (B)

Let $y=\frac{\tan x}{\tan 3x}$
$\Rightarrow y=\frac{\tan x}{\left(\frac{3\tan x-{\tan }^{3}x}{1-3{\tan }^{2}x}\right)}=\frac{1-3{\tan }^{2}x}{3-{\tan }^{2}x}$
$\Rightarrow 3y-y{\tan }^{2}x=1-3{\tan }^{2}x$
$\Rightarrow \left(y-3\right){\tan }^{2}x+\left(1-3y\right)=0$
$\Rightarrow {\tan }^{2}x=\frac{3y-1}{y-3}$
For tan x to be real $\frac{3y-1}{y-3}\ge 0$
$\Rightarrow \left(3y-1\right)\left(y-3\right)\ge 0$ and $y\ne 3$
$\Rightarrow y\le \frac{1}{3}$ or $y>3$