Question

If $-\frac{\pi }{4}<x<\frac{\pi }{4}$ and $1+\left|\tan x\right|+\left|{\tan }^{2}x\right|+...+\infty =\frac{\sqrt{3}(\sqrt{3}+1)}{2}$. Find $x$.

Answer 1

As $-\frac{\pi }{4}<x<\frac{\pi }{4}$

$\therefore |\tan x|<1$

So it's a sum of infinite $G.P$ whose common ratio is $|\tan x|$.

$1+|\tan x|+|{\tan }^{2}x|+...+\infty =\frac{1}{1-|\tan x|}=\frac{\sqrt{3}(\sqrt{3}+1)}{2}$

Solving it we get, $|\tan x|=\frac{1}{\sqrt{3}}\to \tan x=\frac{1}{\sqrt{3}}$ and $\tan x=-\frac{1}{\sqrt{3}}$

$\therefore x=\frac{\pi }{6}$ and $x=-\frac{\pi }{6}$