ban1ka1u

2022-07-07

I know
$g\left(x\right)=\mathrm{arctan}\left(x\right)+\mathrm{arctan}\left(y\right)=\mathrm{arctan}\left(\frac{x+y}{1-xy}\right)$

torpa6d

Expert

Fix, as usual:
$-\frac{\pi }{2}<\gamma =\mathrm{arctan}\left(t\right)<\frac{\pi }{2}$
now we have:
$\mathrm{tan}\left(\gamma \right)=\mathrm{tan}\left(\alpha +\beta \right)=\frac{x+y}{1-xy}=t$
and, if $xy>1$ we have the two cases (x and y have the same sign):
$x>0,y>0\to t<0\to \gamma <0\to \alpha +\beta =\gamma +\pi$
$x<0,y<0\to t>0\to \gamma >0\to \alpha +\beta =\gamma -\pi$

Lucia Grimes

Expert

I can prove that if $|xy|<1$, that
1)
$-\frac{\pi }{2}<\mathrm{arctan}\left(x\right)+\mathrm{arctan}\left(y\right)<\frac{\pi }{2}$
2)
$\mathrm{arctan}\left(x\right)+\mathrm{arctan}\left(y\right)=\mathrm{arctan}\left(\frac{x+y}{1-xy}\right)$

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