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)

$\frac{1}{3}$ and 3

Answer 2

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 $y >3$