Question

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

• A. $Ï•$
• B. { $\frac{Ï€}{4}$ }
• C. { $nÏ€+\frac{Ï€}{4},n=1,2,3,...$ }
• D. { $2nÏ€+\frac{Ï€}{4},n=1,2,3,...$ }

Answer 1

Answer: (A)

We have, $\frac{\tan 3x−\tan 2x}{1+\tan 3x\tan 2x}=1$
$⇒\tan (3x−2x)=1$
$⇒\tan x=1$
$⇒x=nπ+\frac{π}{4}$
But for this value of $x$, we have
$\tan 2x=\tan \left(2nπ+\frac{π}{2}\right)=∞$
which does not satisfy the given equation as it reduces to an indeterminate form.
$∴x=ϕ$