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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:
how do I apply this information to already found $x$ ?
or the domain of whole function is $x \in ]-1;1[$ ?
$\endgroup$2 Answers
$\begingroup$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:
$$]-1,1[.$$
$\endgroup$ $\begingroup$$]-1,1[\,\cap\Bigl(\dfrac\pi2+\mathbf Z \pi\Bigl)=\varnothing$.
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