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function is following: $f(x) = \sin^2 (\frac {2+\tan(x)} {\sqrt{1-x^2}})$

now, $\sin$ domain is $\mathbb{R}$ so I just need to work in the brackets.

the denominator can not be $0$ and the square root must be positive or $0$ so:

$1-x^2 > 0 \implies x^2<1 \implies x \in ]-1;1[$

correct me if I'm wrong.

now my problem is the $\tan(x)$ , the domain of tan function is:

enter image description here

how do I apply this information to already found $x$ ?

or the domain of whole function is $x \in ]-1;1[$ ?

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Your method is great!

For the tangent, you can notice that if $x\in]-1,1[$, then $\tan(x)$ is well-defined (you can see that on the domain you have give for $\tan$), so your final domain is:

$$]-1,1[.$$

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$]-1,1[\,\cap\Bigl(\dfrac\pi2+\mathbf Z \pi\Bigl)=\varnothing$.

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  • $\begingroup$ I don't understand that, $x$ can be $0.5$ for example and $tan(0.5)$ exists $\endgroup$ – Leonardo Jan 31 '17 at 16:17
  • $\begingroup$ This means that, on $\,]-1,1[$, all points are in the domain of $\tan x$. $\endgroup$ – Bernard Jan 31 '17 at 16:19
  • $\begingroup$ a yea, sorry, got it. $\endgroup$ – Leonardo Jan 31 '17 at 16:20
  • $\begingroup$ Maybe I was a little elliptic! $\endgroup$ – Bernard Jan 31 '17 at 16:21

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