In hyperbolic trigonomery, we come across the identity

$\cosh^2 u = \sinh^2 u + 1.$

Recall that from classical trigonometry we also have

$\sec^2 \phi = \tan^2 \phi + 1.$

They look curiously similar and is what would motivate the Gudermannian which seeks to connect classical trigonometry with hyperbolic trigonometry. In fact, the image of $$f(\phi) = (\sec \phi, \tan \phi)$$ is also the hyperbola $$x^2-y^2=1$$.

This motivates the following. Given any point on the right half of the unit circle, $$(\cos \theta, \sin \theta)$$, we could extend it to some point on the right branch of the hyperbola, $$(\cosh u, \sinh u)$$. This in turn gives us an angle $$\phi$$ such that $$(\sec \phi, \tan \phi) = (\cosh u, \sinh u)$$.

As such, we could relate $$u, \theta, \phi$$ by

$\tan \theta = \tanh u = \sin \phi.$

## Derivatives

We could compute that

\begin{align*} \frac{du}{d\theta} &= \sec(2\theta), &\quad \frac{d\theta}{du} &= \text{sech}(2u).\\ \frac{du}{d\phi} &= \sec(\phi), &\quad \frac{d\phi}{du} &= \text{sech}(u).\\ \end{align*}

which seems highly unobvious. We can show that $$\frac{du}{d\theta} = \sec(2\theta)$$ geometrically. First consider the picture below.

We have used the fact the the area of the hyperbolic sector is one half of the hyperbolic angle (similar to how the area of a circular sector in a unit circle is one half of the angle). Approximating the purple area $$du / 2$$ with a circular sector of radius $$\cosh^2 u + \sinh^2 u$$ and angle $$d\theta$$, we get

\begin{align*} \frac{1}{2} \, du &\approx \frac{1}{2}(\cosh^2 u + \sinh^2 u) \, d\theta \\ &= \frac{1}{2}(\sec^2 \phi + \tan^2 \phi) \, d\theta \\ &= \frac{1}{2 \cos^2 \phi}(1 + \sin^2 \phi) \, d\theta \\ &= \frac{1}{2}\frac{1 + \tan^2 \theta}{1-\tan^2 \theta} \, d\theta \\ &= \frac{1}{2} \sec(2\theta) \, d\theta. \end{align*}

## Acknowledgements

Much of this post is inspired from an article by J. M. H. Peters.