Question

$\frac{tan3x}{tanx}$ never lies between

• A. -1/3 and 0
• B. 1/3 and 3
• C. -1/3 and 1/3
• D. 1 and 3

Answer 1

The correct option is B

1/3 and 3

Let y = $\frac{tan3x}{tanx}$ $\Rightarrow$ y = $\frac{3tanx-\left(ta{n}^{3}x\right)}{(1-3ta{n}^{3}x)tanx}$

$\Rightarrow$ $ta{n}^{3}x(3y-1)=y=3$

$\Rightarrow$ $ta{n}^{2}x=\frac{y-3}{3y-1}$

Now $ta{n}^{2}x\ge 0forallx\therefore \frac{y-3}{3y-1}\ge 0$

$\Rightarrow$ (y-3)(3y-1) $\ge$ 0

$\Rightarrow$ y $\le \frac{1}{3}$ or y $\ge$ 3

$\Rightarrow$ y never lies between $\frac{1}{3}$ and 3