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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.

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3 Answers 3

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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 $|y|>y$?

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  • $\begingroup$ 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. $\endgroup$
    – Kanchi
    Jun 23, 2019 at 3:16
  • $\begingroup$ @Kanchi: Yes, that's what I would do. $\endgroup$ Jun 23, 2019 at 8:17
  • $\begingroup$ So I should check for the domain where tanx is >0 in the tanx graph. Ok thank you $\endgroup$
    – Kanchi
    Jun 23, 2019 at 15:00
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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.

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It is $$\frac{x}{\pi}+\frac{1}{2}\notin \mathbb{Z}$$ and $$-\frac{\pi}{2}<x-2\pi n<0$$ or $$\frac{x}{\pi}+\frac{1}{2}\notin \mathbb{Z}$$ and $$\frac{\pi}{2}<x-2\pi n<\pi$$ wher5e $n$ is an integer.

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  • $\begingroup$ This helped me after I checked graph of tanx.Thanks. $\endgroup$
    – Kanchi
    Jun 23, 2019 at 3:43

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