Question

The general solution of the equation $\frac{\tan 3x-\tan 2x}{1+\tan 3x\tan 2x}=1$ is

• A. $\{x:x=n\pi +\frac{\pi }{4},n\in N\}$
• B. $\left\{\frac{\pi }{4}\right\}$
• C. $\{x:x=2n\pi +\frac{\pi }{4},n\in N\}$
• D. $\{x:x=n\pi +(-1{)}^n\frac{\pi }{4},n\in N\}$

Answer 1

The correct option is A $\{x:x=n\pi +\frac{\pi }{4},n\in N\}$

$\frac{\tan 3x-\tan 2x}{1+\tan 3x\tan 2x}=1$
$\Rightarrow \tan (3x-2x)=\tan \frac{\pi }{4}$
$\Rightarrow \tan x=\tan \frac{\pi }{4}$
$\Rightarrow x=n\pi +\frac{\pi }{4}$