Question: Let $\triangle ABC$ and $E, D$ on $[AB]$ and $[AC]$ s.t. $BEDC$ is inscribable. Let $P\in [BD], Q\in [CE]$ s.t. $AEPC$ and $ADQB$ are also inscribable. Show that $AP=AQ$.

I draw the points and I considered $R$, the point of intersections of the two circles, as shown in the following figure.

[![enter image description here](https://i.stack.imgur.com/ozLyQ.jpg)](https://i.stack.imgur.com/ozLyQ.jpg)

Answer: 
Near $0$ you need to expand all terms (into the Taylor series) and cancel $x$:
$$
\frac{\text{e}^x-1}{\sin(x)}=\frac{x+\frac{x^2}{2}+\frac{x^3}{6}+\dots}{x-\frac{x^3}{6}+\dots}=\frac{1+\frac{x}{2}+\frac{x^2}{6}+\dots}{1-\frac{x^2}{6}+\dots}
$$
This formula can be numerically evaluated without problems. If you need a code evaluating this function for any value, then you need this expansion for $x\in[-\varepsilon,\varepsilon]$, where $\varepsilon$ can be explicitly determined (for the required accuracy), and the original formula for $x$ outside of this interval.
*Similar to @Vítězslav Štembera's answer*

Better than Taylor series are Padé approximants : for example
$$\frac{e^x-1}{\sin(x)}=\frac {1+\frac{3 }{14}x+\frac{11 }{84}x^2 } {1-\frac{2 }{7}x-\frac{5 }{84}x^2 }$$ gives an asymptotic error of $\frac{53 }{10080}x^5$

$$\frac{e^x-x}{\sin(x)}=\frac 1 x\, \frac{1-\frac{1}{4}x+\frac{41 }{80}x^2 } {1-\frac{1}{4}x-\frac{37 }{240}x^2 }$$