# Solve 1/(tan2x-tanx)-1/(cot2x-cotx)=1?

Nov 10, 2017

$\frac{1}{\tan 2 x - \tan x} - \frac{1}{\cot 2 x - \cot x} = 1$

$\implies \frac{1}{\tan 2 x - \tan x} - \frac{1}{\frac{1}{\tan 2 x} - \frac{1}{\tan} x} = 1$

$\implies \frac{1}{\tan 2 x - \tan x} + \frac{1}{\frac{1}{\tan x} - \frac{1}{\tan 2 x}} = 1$

$\implies \frac{1}{\tan 2 x - \tan x} + \frac{\tan x \tan 2 x}{\tan 2 x - \tan x} = 1$

$\implies \frac{1 + \tan x \tan 2 x}{\tan 2 x - \tan x} = 1$

$\implies \frac{1}{\tan} \left(2 x - x\right) = 1$

$\implies \tan \left(x\right) = 1 = \tan \left(\frac{\pi}{4}\right)$

$\implies x = n \pi + \frac{\pi}{4}$

Nov 10, 2017

$x = n \pi + \frac{\pi}{4}$

#### Explanation:

$\tan 2 x - \tan x = \frac{\sin 2 x}{\cos 2 x} - \sin \frac{x}{\cos} x = \frac{\sin 2 x \cos x - \cos 2 x \sin x}{\cos 2 x \cos x}$

= $\sin \frac{2 x - x}{\cos 2 x \cos x} = \sin \frac{x}{\cos 2 x \cos x}$

and $\cot 2 x - \cot x = \frac{\cos 2 x}{\sin 2 x} - \cos \frac{x}{\sin} x = \frac{\sin x \cos 2 x - \cos x \sin 2 x}{\sin 2 x \sin x}$

= $\sin \frac{x - 2 x}{\sin 2 x \sin x} = - \sin \frac{x}{\sin 2 x \sin x}$

Hence $\frac{1}{\tan 2 x - \tan x} - \frac{1}{\cot 2 x - \cot x} = 1$ can be written as

$\frac{\cos 2 x \cos x}{\sin} x + \frac{\sin 2 x \sin x}{\sin} x = 1$

or $\frac{\cos 2 x \cos x + \sin 2 x \sin x}{\sin} x = 1$

or $\cos \frac{2 x - x}{\sin} x = 1$

or $\cos \frac{x}{\sin} x = 1$ i.e. $\cot x = 1 = \cot \left(\frac{\pi}{4}\right)$

Hence $x = n \pi + \frac{\pi}{4}$