Trying to find the inverse of $$ 2*\arcsin\left(\frac{1}{1+x}\right) $$
My approach: $$ 2*\arcsin\left(\frac{1}{1+x}\right) = y \iff \arcsin\left(\frac{1}{1+x}\right) = \frac{y}{2}$$ $$ \iff \frac{1}{1+x} = \sin\left(\frac{y}{2}\right) \iff 1+x= \frac{1}{\sin\left(\frac{y}{2}\right)} \iff x = \frac{1}{\sin\left(\frac{y}{2}\right)} - 1 $$
For domain and range of the inverse I have $$\mathbf {D} = \{x \in \Bbb {R}: -\pi \le x \le \pi\} $$ $$\mathbf {R} = \{y \in \Bbb {R}: -2 \le x \le 0, x\ne1\} $$
Checking the inverse and plot in Wolframalpha, it doesn't look at all like this. Is the problem here the multiplication by $(1+x)$ without knowing whether it is positive or negative? How to do this?
You are right and the problem you seem to have interpreting your answer is the following fact: consider the function $$f(x)=\frac{1}{1+x}$$ You have $f(-2)=-1$, $f(0)=1$ and $|f(x)|\gt 1$ for $-2\lt x\lt 0$ hence arcosin $f(x)$ is not defined in this domain despite your answer-function it is (actually the domain of your answer is all $\Bbb R$ excepting the points in which $\sin \frac y2=0$)