I was going through a textbook and came across this expression.

Question

How is this transition possible:
$$e^u = \tan(\pi/4 + \theta/2) \tag{given}$$
$${\cfrac{e^{u/2}}{e^{-u/2}}} =\cfrac{1+\tan(\theta/2)}{1-\tan(\theta/2)} $$

Answer 1

You have: $$\frac{e^\frac{u}{2}}{e^\frac{-u}{2}}=e^u$$ and by usual trigonometry formulas: $$\tan\left( \frac{\pi}{4}+\theta \right)=\frac{\tan\left( \frac{\pi}{4} \right)+\tan(\theta)}{1-\tan\left( \frac{\pi}{4}\right) \tan(\theta)}$$ as $\tan\left( \frac{\pi}{4} \right)=1$ you obtain: $$\tan\left( \frac{\pi}{4}+\theta \right)=\frac{1+\tan(\theta)}{1- \tan(\theta)}$$

Answer 2

$e^u=\tan(\frac\pi4+\frac\theta2) = \frac{1+\tan(\frac\theta2)}{1-\tan(\frac\theta2)}$

$e^u =\dfrac{1+\tan(\frac\theta2)}{1-\tan(\frac\theta2)}$

$e^{\frac u2+\frac u2} = \dfrac{1+\tan(\frac\theta2)}{1-\tan(\frac\theta2)}$

$\dfrac{e^{\frac u2}}{e^{-\frac u2}} =\dfrac{1+\tan(\frac\theta2)}{1-\tan(\frac\theta2)} $