Some Gud Trigonometry

$g,h=i,-i$ or $g,h=-i,i$ , so that $i^2+(-i)^2=-2$

To prove this, $H(G(x))=$

$2ArcTan\left(e^{-i(2ArcTan(e^{ix})-\frac{\pi}{2})}\right)-\dfrac{\pi}{2}$

$2ArcTan\left(ie^{-i(2ArcTan(e^{ix}))}\right)-\dfrac{\pi}{2}$

$2ArcTan\left(ie^{-i\left(2\left(\frac{1}{2}iLog\left(\frac{1-ie^{ix}}{1+ie^{ix}} \right) \right) \right) }\right)-\dfrac{\pi}{2}$

$2ArcTan\left(ie^{Log\left(\frac{1-ie^{ix}}{1+ie^{ix}} \right) }\right)-\dfrac{\pi}{2}$

$2ArcTan\left( i \dfrac{1-ie^{ix}}{1+ie^{ix}} \right)-\dfrac{\pi}{2}$

$2iArcTanh\left(\dfrac{1-ie^{ix}}{1+ie^{ix}} \right)-\dfrac{\pi}{2}$

$2iArcTanh\left(\dfrac{1-e^{i(x+\frac{\pi}{2})}}{1+e^{i(x+\frac{\pi}{2})}} \right)-\dfrac{\pi}{2}$

$2iArcTanh\left(Tanh\left( -\frac{1}{2}i\left(x+\frac{\pi}{2}\right) \right) \right)-\dfrac{\pi}{2}$

$2i \left( -\frac{1}{2}i\left(x+\frac{\pi}{2} \right) \right)-\dfrac{\pi}{2}$

$x$

This function (notice, no $i$ ):

$2ArcTan\left(e^x \right)-\dfrac{\pi}{2}$

is the Gudermannian Function $gd(x)$ , with the property that

$gd^{-1}(ix)=-igd(x)$

(negative exponent -1 indicates inverse function) and connects oridinary trigonometric functions with hyperbolic trigonometric functions, as well as their inverses, such as

$Tan(gd(x))=Sinh(x)$
$Tanh(gd^{-1}(x))=Sin(x)$
$gd(Tanh^{-1}(x))=Sin^{-1}(x)$
$gd^{-1}(Tan^{-1}(x))=Sinh^{-1}(x)$

so that, for example, we can rearrange the second to last one to

$gd(x)=Sin^{-1}(Tanh(x)) = ArcSin(Tanh(x))$

for another definition of the Gudermannian function, one of many more possible. We can even see that

$ArcSin(Tanh(x)) = ArcTan(Sinh(x))$

and therefore

$Tan(ArcSin(Tanh(x))=Sinh(x)$

etc