Question

Set of values for which $\frac{\tan 3x-\tan 2x}{1+\tan 3x\tan 2x}=1$ is true is

• A. $\varphi$
• B. $n\pi +\frac{\pi }{4},n\in Z$
• C. $\left(\frac{\pi }{4}\right)$
• D. $2n\pi +\frac{\pi }{4},n\in Z$

Answer 1

The correct option is A $\varphi$

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

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

$\Rightarrow \tan x=1$ provided $\tan 2x\tan 3x\ne -1$

If solve the above equation, we get x to be $n\pi +\frac{\pi }{4}$

But, that will make $\tan 2x$ to be infinite and therefore, this value cannot be accepted.

So, there is no such $x$ satisfying the above equality.