Question

The range of $x$ satisfying $\tan x-{\tan }^{2}x>0$ and $|2\sin x|<1$ is
(where $n\in Z$)

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

Answer 1

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

Given: $\tan x-{\tan }^{2}x>0$ and $|2\sin x|<1$
Now,
$|2\sin x|<1\Rightarrow |\sin x|<\frac{1}{2}\Rightarrow -\frac{\pi }{6}<x<\frac{\pi }{6}$
As the period of $|\sin x|$ is $\pi$, so
$-\frac{\pi }{6}+n\pi <x<\frac{\pi }{6}+n\pi ,n\in Z$
Also,
$\tan x-{\tan }^{2}x>0\Rightarrow \tan x(\tan x-1)<0\Rightarrow 0<\tan x<1\Rightarrow 0<x<\frac{\pi }{4}$
As the period of $\tan x$ is $\pi$, so
$\Rightarrow n\pi <x<n\pi +\frac{\pi }{4},n\in Z$

Therefore,
$x\in (n\pi ,n\pi +\frac{\pi }{6}),n\in Z$