A human way to simplify $ \frac{((\sqrt{a^2 - 1} - a)^2 - 1)^2}{(\sqrt{a^2 - 1} - a)^22 \sqrt{a^2 - 1}} - 2 a $ I end up with simplifying the following fraction when I tried to calculate an integral(*) with the residue theory in complex analysis:
$$
\frac{((\sqrt{a^2 - 1} - a)^2 - 1)^2}{(\sqrt{a^2 - 1} - a)^22 \sqrt{a^2 - 1}} - 2 a
$$
where $a>1$. With Mathematica, I can quickly get
$$
\frac{((\sqrt{a^2 - 1} - a)^2 - 1)^2}{(\sqrt{a^2 - 1} - a)^22 \sqrt{a^2 - 1}} - 2 a=2(\sqrt{a^2 - 1} - a).
$$
Would anybody give a calculation for the simplification in a human way? 

(*)The integral I did is
$$
\int_{-\pi}^\pi\frac{\sin^2 t}{a+\cos t}\ dt
$$
with $a>1$.

Start with
$$\begin{align}\left(\sqrt{a^2-1}-a\right)^2-1&=a^2-1-2a\sqrt{a^2-1}+a^2-1\\
&=\sqrt{a^2-1}\left(2\sqrt{a^2-1}-2a\right)\\
&=2\sqrt{a^2-1}\left(\sqrt{a^2-1}-a\right)\end{align}$$
So you are now down to
$$\frac{\left(2\sqrt{a^2-1}\left(\sqrt{a^2-1}-a\right)\right)^2}{\left(\sqrt{a^2-1}-a\right)^2\cdot2\sqrt{a^2-1}}-2a=2\sqrt{a^2-1}-2a$$
