<s>[INST] inverse of $\arcsin (\frac{x}{x-1})$ determine the inverse of
a) $y=\arcsin \left(\dfrac{x}{x-1}\right)$
b) $y= \dfrac{1-2e^{-x}}{4}$
I learned you the steps for finding the inverse are
1) get it in a form of $x= \dots$
2) change $x$ and $y$ 
a) $y=\arcsin \left(\dfrac{x}{x-1}\right)$
is $y=\arcsin \left(\dfrac{x}{x-1}\right)\quad \iff \quad \sin y = \dfrac{x}{x-1}$? and if so what next?
b) $y= \dfrac{1-2e^{-x}}{4} \quad \iff \quad 4y-1=-2e^{-x} \quad\iff\quad -2y + \frac{1}{2}= e^{-x}$ . but then I'm stuck.
I would really appreciate the help! 
 [/INST] $\sin y(x-1)=x\Rightarrow \sin y x-x=siny \Rightarrow x(\sin y -1)=\sin y$ so $$x=\frac{\sin y}{\sin y -1}\Rightarrow y^{-1}=\frac{\sin x}{\sin x -1}$$
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