Question

The values of $x\in (-\frac{\pi }{2},\frac{\pi }{2})$ for which both the conditions ${\tan }^{2}x-\tan x<0$ and $2|\sin x|<1$ satisfy simultaneously are

• A. $(0,\frac{\pi }{6})$
• B. $(0,\frac{\pi }{4})$
• C. $(-\frac{\pi }{6},0)$
• D. $(-\frac{\pi }{6},\frac{\pi }{6})$

Answer 1

The correct option is A $(0,\frac{\pi }{6})$

Given : $x\in (-\frac{\pi }{2},\frac{\pi }{2})$
${\tan }^{2}x-\tan x<0$
$\Rightarrow \tan x(\tan x-1)<0$
$\Rightarrow \tan x\in (0,1)$
$\Rightarrow x\in (0,\frac{\pi }{4})\cdots \left(1\right)$

and $2|\sin x|<1$
$\Rightarrow |\sin x|<\frac{1}{2}$
$\Rightarrow -\frac{1}{2}<\sin x<\frac{1}{2}$
$\Rightarrow x\in (-\frac{\pi }{6},\frac{\pi }{6})\cdots \left(2\right)$
From $\left(1\right)\cap \left(2\right)$ : $x\in (0,\frac{\pi }{6})$