Question

The number of solution(s) of $\frac{\tan 3x-\tan 2x}{1+\tan 3x\tan 2x}=1$ is

Answer 1

$\frac{\tan 3x-\tan 2x}{1+\tan 3x\tan 2x}=1\Rightarrow \tan (3x-2x)=1\Rightarrow \tan x=1\Rightarrow x=n\pi +\frac{\pi }{4}$
Therefore,
$2x=2n\pi +\frac{\pi }{2},3x=3n\pi +\frac{3\pi }{4}$
So, $\tan 2x=\tan (2n\pi +\frac{\pi }{2})=\tan \left(\frac{\pi }{2}\right)$
As $\tan 2x$ is not defined, so the given equation has no solution.