<s>[INST] Why does $\arctan(\frac{\tan \theta}{2}) \approx \frac{1}{2 - \theta} - \frac{1}{2 + \theta}$ for small $\theta$? In answering this question, I needed to show
that
$\arctan \left( \frac{\tan \theta}{2} \right) \approx \frac{\theta}{2}$
when $\theta$ was small.
So, naturally,
I computed the first few terms of the Taylor series for $\arctan \left( \frac{\tan \theta}{2} \right)$.
Sure enough,
the first three terms are $0 + \frac12 \theta + 0 \theta^2$.
But what I didn't expect was the striking pattern in the next few terms:
$$
\arctan \left( \frac{\tan \theta}{2} \right)
= \frac12 \theta
  + \frac{1}{8} \theta^3
  + \frac{1}{32} \theta^5
  + \frac{11}{1920} \theta^7 + \cdots
$$
It clearly seems to be the case that
\begin{align*}
\arctan \left( \frac{\tan \theta}{2} \right)
&\approx \sum_{n \ge 0} \left(\frac{\theta}{2}\right)^{2n+1} \\
&= \frac{\theta / 2}{1 - \theta^2 / 4} \\
&= \frac{1}{2 - \theta} - \frac{1}{2 + \theta}. \\
\end{align*}
Here is a plot:

Honestly, I have no idea how to explain this, and in particular I'm having trouble assigning any meaning to $\frac{1}{2 - x}$ at all when $x$ is
an angle.
So, my question: is there any intuitive geometric reason why this approximation works?  If not, is there any other concise reason?
Obviously one can arrive at this result via algebraic manipulation (like I did above) but I am interested in a more intuitive explanation.
 [/INST] The approximation you noticed is the Padé approximant of order $(m,n) = (1,2)$ of $f(\theta) = \tan^{-1} \bigl( \frac{1}{2} \tan \theta \bigr)$.  
</s>