# To find the domain of a function

How to find the domain of the function $$f(x)=\frac{1}{\sqrt{|\tan x| -\tan(x)}}?$$

My attempt: since the expression under the root should be greater than zero so $$\vert \tan x \vert - \tan x > 0$$ and $$\vert \tan x \vert > \tan x$$ but I'm stuck here so please help. I thought of taking $$x<0$$ and $$x>0$$ but then I'm not able to bring out the answer.

Instead of dividing according to the sign of $$x$$, give $$\tan x$$ a name such as $$y$$, and ask yourself:

For which $$y$$ do we have $$|y|>y$$?

• So if I put tanx=y and compare |y|>y then I get y>0 for |y|=(-y) , so tanx>0 , is this the way it should be done?Please check. Jun 23, 2019 at 3:16
• @Kanchi: Yes, that's what I would do. Jun 23, 2019 at 8:17
• So I should check for the domain where tanx is >0 in the tanx graph. Ok thank you Jun 23, 2019 at 15:00

Given the function $$f : D_f \to \mathbb{R}$$ of law: $$f(x) := \frac{1}{\sqrt{|\tan x| - \tan x}}\,,$$ its natural domain is thus determinable: $$|\tan x| > \tan x \; \; \; \; \; \; \Leftrightarrow \; \; \; \; \; \; \begin{cases} \tan x < 0 \\ - \tan x > \tan x \end{cases} \; \; \; \cup \; \; \; \begin{cases} \tan x \ge 0 \\ \tan x > \tan x \end{cases}\,.$$ To conclude you.

It is $$\frac{x}{\pi}+\frac{1}{2}\notin \mathbb{Z}$$ and $$-\frac{\pi}{2} or $$\frac{x}{\pi}+\frac{1}{2}\notin \mathbb{Z}$$ and $$\frac{\pi}{2} wher5e $$n$$ is an integer.

• This helped me after I checked graph of tanx.Thanks. Jun 23, 2019 at 3:43