reverse of $f(x) = 1 + \tan(x^2)$

Question

how to do make the inverse of this function, I already have the answer but I cant see how it got there:

this is the answer:

$$\pm \sqrt{-\tan^{-1}(1-x)}$$

Answer 1

Hint

Isolate $y$

$$x=1+\tan(y^2)$$

Also remember that $\tan x$ is an odd function. It means that $-\tan^{-1}(1-x)=\tan^{-1}(x-1)$

Can you finish?

Answer 2

Let $y=f(x)=1+\tan(x^2)$

$\implies\tan(x^2)=y-1\implies x^2=\arctan(y-1)\implies x=\pm\sqrt{\arctan(y-1)}$

$\implies f^{-1}(y)=x=\pm\sqrt{\arctan(y-1)}$