Question

The set of values of $ x $ for which $ \frac{\tan 3x - \tan 2x}{1+ \tan 3x\,\tan2x} = 1 $ is

• A. $ \phi $
• B. { $ \frac{\pi}{4} $ }
• C. { $ n\pi + \frac {\pi}{4} , n = 1,2, 3,... $ }
• D. { $ 2n\pi + \frac {\pi}{4} , n = 1,2, 3,... $ }

Answer 1

Answer: (A)

We have, $\frac{\tan 3 x-\tan 2 x}{1+\tan 3 x \tan 2 x}=1$
$\Rightarrow \tan (3 x-2 x)=1$
$\Rightarrow \tan x=1$
$\Rightarrow x=n \pi+\frac{\pi}{4}$
But for this value of $x$, we have
$\tan 2 x=\tan \left(2 n \pi+\frac{\pi}{2}\right)=\infty$
which does not satisfy the given equation as it reduces to an indeterminate form.
$\therefore x=\phi$