I am trying to show that the parametrisation $$\varphi(u,v) = (u \cos (v), sech (u), \tanh (u))$$ of the unit sphere $S^2$ (the Mercator's Projection) can be given by the change of coordinates $\log \tan(\frac{1}{2}\theta) = u$ and $\varphi = v$ from the parametrisation, $$X(\theta,\varphi) = (\sin\theta \cos \varphi, \sin \theta \sin \varphi, \cos \theta)$$ of $S$, where $X: U \in \mathbb{R}^2 \rightarrow \mathbb{R}^3$ from the open neighbourhood $U = \{(\theta, \varphi) \in \mathbb{R}^3; 0< \theta < \pi,~ 0<\varphi<2\pi\}$.
In an attempt to reverse engineer, I substituted $u= \log \tan(\frac{1}{2}\theta)$ and $v=\varphi$ into $\varphi(u,v) $,
where $\varphi(u,v)$ then becomes $(\log \tan(\frac{1}{2}\theta)\cos v, sech(\log \tan(\frac{1}{2}\theta)), \tanh (\log \tan(\frac{1}{2}\theta)).$ This looked impossible to simplify so I tried expressing the hyperbolic trigonometric functions of $\varphi$ as an expression of complex numbers, so $sech (u) = \frac{2}{e^u+e^{-u}}$ and $\tanh (u) = \frac{e^u-e^u}{e^u+e^{-u}}$, which gives $\varphi(u,v) = (u\cdot \frac{e^{iv}+e^{-iv}}{2}, \frac{2}{e^u+e^{-u}}, \frac{e^u-e^u}{e^u+e^{-u}})$, but I still need to relate them to $X$.
Plugging in my values $u= \log \tan(\frac{1}{2}\theta)$ and $v=\varphi$,
$$\varphi(u,v) = \left(u\cdot \frac{e^{iv}+e^{-iv}}{2}, \frac{2}{e^u+e^{-u}}, \frac{e^u-e^u}{e^u+e^{-u}}\right)$$
$$\varphi(u,v) = \left(u\cdot \frac{e^{i\varphi}+e^{-i\varphi}}{2}, \frac{2}{e^{(\log\tan(\frac{1}{2}\theta)}+e^{\log \tan^{-1}(\frac{1}{2}\theta)}}, \frac{e^{\log \tan(\frac{1}{2}\theta)}-e^{\log \tan^{-1}(\frac{1}{2}\theta)}}{e^{\log \tan(\frac{1}{2}\theta)}+e^{\log \tan^{-1}(\frac{1}{2}\theta)}}\right)$$
$$\varphi(u,v) = \left(\log \tan(\frac{1}{2}\theta)\cdot \frac{e^{i\varphi}+e^{-i\varphi}}{2}, \frac{2}{\tan(\frac{1}{2}\theta)+\tan^{-1}(\frac{1}{2}\theta)}, \frac{\tan(\frac{1}{2}\theta)-\tan^{-1}(\frac{1}{2}\theta)}{\tan(\frac{1}{2}\theta)+ \tan^{-1}(\frac{1}{2}\theta)}\right)$$
$$\varphi(u,v) = \left(\frac{e^{i\theta}-e^{-i\theta}}{e^{i\theta}+e^{-i\theta}}\cdot \cos\varphi, \sin\theta, 1-\cos\theta\right)$$
Unfortunately this already looks wrong, and I cannot get it to the parametrisation $X$.
I've also tried rearranging for $\theta$ to get $e^u = \tan \left(\frac{1}{2}\theta \right)$, so $\theta = 2\arctan (e^u)$, but I'm finding it very difficult to simiplify the expressions in any way.
Any help would be appreciated!
If we start by putting $t=\tan{\frac{1}{2}\theta}$, then $$ (\sin{\theta}\cos{\phi},\sin{\theta}\sin{\phi},\cos{\theta}) = \left( \frac{2t}{1+t^2} \cos{\phi} , \frac{2t}{1+t^2} \sin{\phi} , \frac{1-t^2}{1+t^2} \right) $$ by using the half-angle formulae. Now writing $t=e^u$, we have $$ \left( \frac{2t}{1+t^2} \cos{\phi} , \frac{2t}{1+t^2} \sin{\phi} , \frac{1-t^2}{1+t^2} \right) = \left( \frac{2e^u}{1+e^{2u}} \cos{\phi} , \frac{2e^u}{1+e^{2u}} \sin{\phi} , \frac{1-e^{2u}}{1+e^{2u}} \right) \\ = \left( \frac{2}{e^{-u}+e^{u}} \cos{\phi} , \frac{2}{e^{-u}+e^{u}} \sin{\phi} , \frac{e^{-u}-e^{u}}{e^{-u}+e^{u}} \right) \\ = ( \operatorname{sech}{u} \cos{v} , \operatorname{sech}{u} \sin{v} , -\tanh{u} ) $$ since $\phi=v$. Therefore you probably want $u=\log{\cot{\frac{1}{2}\theta}}$.