Question

If $\tan x-{\tan }^{2}x>0$ and $|2\sin x|<1$ then the range of $x$ for which both conditions hold good is (where $n\in Z$)

• A. $\bigcup _{n\in Z}(n\pi ,n\pi +\frac{\pi }{6})$
• B. $\bigcup _{n\in Z}(n\pi -\frac{\pi }{4},n\pi +\frac{\pi }{6})$
• C. $\bigcup _{n\in Z}(n\pi -\frac{\pi }{6},n\pi +\frac{\pi }{6})$
• D. $\bigcup _{n\in Z}(n\pi ,n\pi +\frac{\pi }{4})$

Answer 1

The correct option is A $\bigcup _{n\in Z}(n\pi ,n\pi +\frac{\pi }{6})$

$\tan x-{\tan }^{2}x>0$
$\Rightarrow \tan x(\tan x-1)<0$
$\Rightarrow 0<\tan x<1$
$\Rightarrow 0<x<\frac{\pi }{4}$
$\Rightarrow n\pi <x<n\pi +\frac{\pi }{4},n\in Z$
$|\sin x|<\frac{1}{2}$
$\Rightarrow -\frac{\pi }{6}<x<\frac{\pi }{6}\because \text{period of}\sin x\text{is}\pi \Rightarrow -\frac{\pi }{6}+n\pi <x<\frac{\pi }{6}+n\pi ,n\in Z$
$\therefore \text{Solution set for both the given inequalities is}n\pi <x<n\pi +\frac{\pi }{6}.$