Question

The general solution of $\frac{\tan 2x-\tan x}{1+\tan x\tan 2x}=1$ is

• A. $n\pi +\frac{\pi }{4},\forall n\in Z$
• B. $n\pi \pm \frac{\pi }{4},\forall n\in Z$
• C. $\varphi$
• D. $n\pi +\frac{\pi }{6},\forall n\in Z$

Answer 1

The correct option is A $n\pi +\frac{\pi }{4},\forall n\in Z$

Formula:$\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$

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

$\Rightarrow \tan x=1$

$\therefore x\in n\pi +\frac{\pi }{4}$, $n\in z$.