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Q.
The value of $\frac{\tan \, x}{\tan \, 3x}$ wherever defined never lies between
Trigonometric Functions
Solution:
Let $y = \frac{\tan x}{\tan3x}$
$ \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 $\frac{3y-1}{y-3} \ge0$
$ \Rightarrow \left(3y -1\right)\left(y-3\right) \ge 0$ and $ y \ne3$
$ \Rightarrow y \le \frac{1}{3}$ or $ y >3$