duality of Gudermannian and its inverse function

There are a lot of formulae concerning the Gudermannian function and its inverse function containing a hyperbolic function or a trigonometric function or both, such that if we change functions of one kind to the corresponding functions of the other kind, then the new formula also is true.

Some exemples:

$\text{gd}x={\displaystyle {\int }_0^x}{\displaystyle \frac{dt}{\cosh t}},{\text{gd}}^{-1}x={\displaystyle {\int }_0^x}{\displaystyle \frac{dt}{\cos t}}$(1)

${\displaystyle \frac{d}{dx}}\text{gd}x={\displaystyle \frac{1}{\cosh x}},{\displaystyle \frac{d}{dx}}{\text{gd}}^{-1}x={\displaystyle \frac{1}{\cos x}}$(2)

$\tan (\text{gd}x)=\sinh x,\tanh ({\text{gd}}^{-1}x)=\sin x$(3)

$\sin (\text{gd}x)=\tanh x,\sinh ({\text{gd}}^{-1}x)=\tan x$(4)

$\tan {\displaystyle \frac{\text{gd}x}{2}}=\tanh {\displaystyle \frac{x}{2}},\tanh {\displaystyle \frac{{\text{gd}}^{-1}x}{2}}=\tan {\displaystyle \frac{x}{2}}$(5)

For proving (5) we can check that

$$\frac{d}{dx}[2\arctan (\tanh \frac{x}{2})]=\frac{1}{\cosh x},$$

and since both the expression in the brackets and the http://planetmath.org/node/11997Gudermannian vanish in the origin, we have

$$\text{gd}x\equiv \mathrm{\hspace{0.33em}2}\arctan (\tanh \frac{x}{2}).$$

This equation implies (5).

The duality (http://planetmath.org/DualityInMathematics) of the formula pairs may be explained by the equality

$\text{gd}ix=i{\text{gd}}^{-1}x.$(6)