Algebra problem with isomorphism Let $G=(-1,1)$ and the binary operation  such that
$x*y=\frac{x\sqrt{1-y^2}+y\sqrt{1-x^2}}{\sqrt{1-(xy)^2+2xy\sqrt{(1-x^2)(1-y^2)}}}$.
Prove that (G,*) it is a isomorph group with the group (R,+).
I tried to find the bijective function $f:G\to R$ such that  $f(x*y)=f(x)+f(y)$   but I could not.

Let $x=\sin\alpha$, $y=\sin\beta$, so $\alpha,\beta\in(-\pi/2,\pi/2)$. (always do such substitution when there is a $\sqrt{1-x^2}$)
Then $x*y = \frac{\sin\alpha\cos\beta+\cos\alpha\sin\beta}{\sqrt{1 - \sin^2\alpha\sin^2\beta + 2\sin\alpha\sin\beta\cos\alpha\cos\beta}} = \frac{\sin(\alpha+\beta)}{\sqrt{\cos^2\alpha\cos^2\beta + \sin^2(\alpha+\beta)}} = \frac{t}{\sqrt{1+t^2}} = \tanh\mathrm{asinh}\,t = g(t) $ where $t = \frac{\sin(\alpha+\beta)}{\cos\alpha\cos\beta} = \tan\alpha+\tan\beta = \frac{x}{\sqrt{1-x^2}} + \frac{y}{\sqrt{1-y^2}} = f(x)+f(y) $.
(I have replaced $1$ in the denominator by $(\sin^2\alpha+\cos^2\alpha)(\sin^2\beta+\cos^2\beta)$ because we need something symmetric with respect to $\alpha\leftrightarrow\beta$, and we need to get rid of that complicated term with the factor $2$; we did it by including it into $\sin^2(\alpha+\beta)$.)
So $x*y=g(f(x)+f(y))$, and since in fact $g(f(x))=x$, $f$ and $g$ indeed give the desired isomorphism.
