Question

Solve $\frac{\tan 3x-\tan 2x}{1+\tan 3x\tan 2x}=1$

Answer 1

Given that

$\frac{\tan 3x-\tan 2x}{1+\tan 3x\tan 2x}=1$

The above equation is in the form of $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$

$\Rightarrow \tan (3x-2x)=1$

$\Rightarrow (3x-2x)={\tan }^{-1}\left(1\right)$

$\Rightarrow x=\frac{\pi }{4}$