The map \[ x\mapsto\frac{1-x^2}{1+x^2}+i\,\frac{2x}{1+x^2}=\frac{1+ix}{1-ix},\qquad x\in\RR, \] is a bijection from \(\RR\) onto \(S^1\setminus\{-1\}\). Substituting \(x=\tan t\) for \(t\in\qty{-\frac{\pi}{2},\frac{\pi}{2}}\) gives \[ F(t)\coloneqq\frac{1-x^2}{1+x^2}+i\,\frac{2x}{1+x^2}=\cos 2t+i\sin 2t=e^{2it}. \]
Since \((1+ix)(1-ix)=1+x^2\), we have \(\frac{1+ix}{1-ix}=\frac{(1+ix)^2}{1+x^2}=\frac{1-x^2+2ix}{1+x^2}\), and its modulus is \(\abs{1+ix}/\abs{1-ix}=1\). For \(t\in(-\pi/2,\pi/2)\) and \(x=\tan t\), \(1+x^2=\sec^2t\), so \(\frac{1-x^2}{1+x^2}=\cos^2t-\sin^2t=\cos2t\) and \(\frac{2x}{1+x^2}=2\sin t\cos t=\sin2t\). As \(t\) runs over \((-\pi/2,\pi/2)\), \(x=\tan t\) runs bijectively over \(\RR\) and \(2t\) runs over \((-\pi,\pi)\), so \(e^{2it}\) runs bijectively over \(S^1\setminus\{-1\}\).