Domain of $g(x)=\frac{1}{1-\tan x}$

Question

What is the domain of $g(x)=\frac{1}{1-\tan x}$

I tried it and got this. But I'm not really sure if it is right. Is that gonna be like this ? $(\mathbb{R}, \frac{\pi}{4})$

Answer 1

It is $\mathbb R \setminus \cup_{\{n \in \mathbb Z\} }(\{n\pi +\pi /2\} \cup \{n\pi +\pi /4\})$

Answer 2

Any fraction is not defined if its denominator is zero. Hence you must exclude all the points where $\tan x=1$,which are $x\in\{k\pi+\arctan 1 | k\in \mathbb{Z}\}=\{k\pi + \frac{\pi}{4}| k\in \mathbb{Z}\}$.
However, the $\tan$ function itself is undefined if $x\in\{k\pi+\frac{\pi}{2} | k\in \mathbb{Z}\}$,so we may conclude that domain of $g$ is $R\setminus \left(\{k\pi+\frac{\pi}{2} | k\in \mathbb{Z}\}\cup\{k\pi + \frac{\pi}{4}| k\in \mathbb{Z}\}\right) $.