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 $a>1$.

The substitution $a=\cosh \tau$ is convenient here, since then $$\sqrt{a^2-1}-a=\sinh \tau-\cosh \tau = -e^{-\tau }.$$ Hence the first term of the expression in the question becomes 
$$\frac{(e^{-2\tau} - 1)^2}{e^{-2\tau} (e^\tau-e^{-\tau})}=\frac{(e^{-\tau} - e^{\tau})^2}{(e^\tau-e^{-\tau})}=e^{\tau}-e^{-\tau}=2\sinh\tau=2\sqrt{a^2-1}$$
in agreement with Mathematica.
