1
$\begingroup$

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 $x \in ]-1;1[$ ?

2 Answers 2

1

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:

$$]-1,1[.$$

1

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

  • 0
    I don't understand that, $x$ can be $0.5$ for example and $tan(0.5)$ exists – 2017-01-31
  • 0
    This means that, on $\,]-1,1[$, all points are in the domain of $\tan x$. – 2017-01-31
  • 0
    a yea, sorry, got it. – 2017-01-31
  • 0
    Maybe I was a little elliptic! – 2017-01-31