<s>[INST] How to prove that $ \arctan x + 2 \arctan(\sqrt{1 + x^2} - x) = \frac{\pi}{2} $ I want to prove that: $$\arctan x + 2 \arctan(\sqrt{1 + x^2} - x) = \frac{\pi}{2}, \forall x \in \mathbb{R} $$ 
I know that $$ \arctan x + arctan \frac{1}{x} = \frac{\pi}{2} $$
But that doesn't seem to be helping. How should I proceed?
 [/INST] You may set $x=\tan\theta$ (for some $\theta\in\left(-\pi/2,\pi/2\right)$) and check that
$$2\arctan\left(\frac{1-\sin\theta}{\cos\theta}\right) = \frac{\pi}{2}-\theta $$
holds, or that
$$ 2\arctan\left(\frac{1-\cos\theta}{\sin\theta}\right)=\theta $$
holds for any $\theta\in(0,\pi)$. Since $1-\cos\theta=2\sin^2\frac{\theta}{2}$ and $\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$ this is equivalent to
$$ 2 \arctan\tan\frac{\theta}{2} = \theta $$
for any $\theta\in(0,\pi)$, which is fairly obvious.
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