Question

Solve the inequality : ${\tan }^{2}x-(\sqrt{3}+1)\tan x+\sqrt{3}<0$.

• A. $n\pi +\frac{\pi }{3}<x<n\pi +\frac{\pi }{2},n\in I$
• B. $n\pi +\frac{\pi }{4}<x<n\pi +\frac{\pi }{3},n\in I$
• C. $n\pi +\frac{\pi }{6}<x<n\pi +\frac{\pi }{3},n\in I$
• D. $n\pi +\frac{\pi }{4}<x<n\pi +\frac{\pi }{2},n\in I$

Answer 1

The correct option is B $n\pi +\frac{\pi }{4}<x<n\pi +\frac{\pi }{3},n\in I$

${\tan }^{2}x-(\sqrt{3}+1)\tan x+\sqrt{3}<0$

$\Rightarrow {\tan }^{2}x-\sqrt{3}\tan x-\tan x+\sqrt{3}<0$

$\Rightarrow \tan x(\tan x-\sqrt{3})-(\tan x-\sqrt{3})<0$

$\Rightarrow (\tan x-1)(\tan x-\sqrt{3})<0$

$\Rightarrow 1<\tan x<\sqrt{3}$

$\Rightarrow n\pi +\frac{\pi }{4}<x<n\pi +\frac{\pi }{3}$