How to simplify $\tan{\arcsin{\frac{y}{R}}/2}$? I have verified with Mathematica that, for $R>0, y \in \mathbb{R}$:
$$ \tan{\frac{\arcsin{\frac{y}{R}}}{2}} = \frac{R - \sqrt{R^2 -y ^2}}{y}$$
using 
Assuming[Element[y, Reals] && R > 0, 
 FullSimplify[TrigToExp[Tan[ArcSin[y/R]/2]]]]

How can I prove this withouse messing with complex exponentials and logarithms?
 A: Create a right triangle with sides $x,y$, and hypotenuse $R$. In the diagram below, $\phi = \arcsin(y/R) / 2$, So you are looking to find $\tan \phi$ in terms of $R$ and $y$. 
For this purpose, you may find the angle bisector theorem useful, as it tells you how the two sections of $y$ are related:
$$\tan \phi = \frac{y_1}{x}=\frac{y_2}{R}$$
Or that:
$$y_1 = \left(\frac{R}{x}+1\right)^{-1}y \quad\longrightarrow\quad 
\tan \phi=\frac{y}{\sqrt{R^2-y^2}+R}$$
Now multiply the denominator by it's conjugate to obtain your result. 
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$
A: Let  $\displaystyle\arcsin\frac yr=\phi\implies \sin\phi=\frac yR\ \ \ \ (1)$
and based on the definition of principal value, $\displaystyle-\frac\pi2\le\phi\le\frac\pi2\ \ \ \  (2)$
So, we need to find $\displaystyle\tan\frac\phi2$
Using Weierstrass substitution, $\displaystyle\sin\phi=\frac{2\tan\frac\phi2}{1+\tan^2\frac\phi2}$
So, using  $\displaystyle(1),\frac{2\tan\frac\phi2}{1+\tan^2\frac\phi2}=\frac yR\implies y\tan^2\frac\phi2-2R\tan\frac\phi2+y=0\  \ \ \ (3)$
Solving the Quadratic Equation for $\displaystyle\tan\frac\phi2=\frac{R\pm\sqrt{R^2-y^2}}y$
Now using $\displaystyle(2), -\frac\pi4\le\frac\phi2\le\frac\pi4\implies -1\le\tan\frac\phi2\le1$
Observe that for real $\displaystyle\phi, R^2\ge y^2\implies R\ge |y|$ as $R>0$
If $\displaystyle R=|y|,\tan\frac\phi2=\pm1$ where both roots are same
Else  $\displaystyle\frac{R+\sqrt{R^2-y^2}}{|y|}> \frac R{|y|}>1,$ hence  should be discarded.

Observe that if $\displaystyle\tan\frac{\phi_1}2,\tan\frac{\phi_2}2$ are the roots of $(3),$
using Vieta's formula, $\displaystyle\tan\frac{\phi_1}2\tan\frac{\phi_2}2=1$
$\displaystyle\implies\tan\frac{\phi_1}2=\frac1{\tan\frac{\phi_2}2}=\cot\frac{\phi_2}2=\tan\left(\frac\pi2-\frac{\phi_2}2\right)=\tan\frac{\pi-\phi_2}2$
$\displaystyle\implies\frac{\phi_1}2=\frac{\pi-\phi_2}2\iff\phi_1=\pi-\phi_2 $
$\displaystyle\implies\sin\phi_1=\sin\left(\pi-\phi_2\right)=\sin\phi_2 $
Also,  $\displaystyle\sin\phi=\frac{2\tan\frac\phi2}{1+\tan^2\frac\phi2}=\frac{2\cot\frac\phi2}{1+\cot^2\frac\phi2}$ (multiplying the numerator & the denominator by $\displaystyle\cot^2\frac\phi2$)
