<s>[INST] inverse of a funtion I'm trying to calculate the inverse of the function:
$$y= \tan \left( \cos^{-1}  \frac{x}{x+2} \right)  $$
In my opinion it is $y=  \frac{2 \cos ( \tan^{-1} x )}{1-\cos ( \tan^{-1} x )}  $ but in my book it is reported :
$f(x)\begin{cases} -2\left( \frac{\sqrt{1+x^2}-1}{x^2} \right) &  x < 0\\
-1 & x = 0\\2 \left( \frac{\sqrt{1+x^2}+1}{x^2} \right) & x>0 \end{cases}$
Can someone explain me the passages?
 [/INST] Hint
$$y= \tan\big[ \arccos(f(x)) \big]=\frac{\sqrt{1-f(x)^2}}{f(x)}$$  could help you.
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