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?
 A: 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.
A: We have 
\begin{eqnarray*}
\tan^{-1}(A) + \tan^{-1}(B) + \tan^{-1}(C)  = \tan^{-1}   \left(  \frac{A+B+C-ABC}{1-AB-BC-CA}  \right). 
\end{eqnarray*}
If the RHS is to give $ \pi/2$ then we require the denominator to be zero, so
\begin{eqnarray*}
1-2x( \sqrt{1+x^2} -x) -(\sqrt{1+x^2} -x)^2=0
\end{eqnarray*}
Which is easily verified.
A: Taking a derivative of the expression and you will get
$$
\newcommand{\Bold}[1]{\mathbf{#1}}\frac{2 \, {\left(\frac{x}{\sqrt{x^{2} + 1}} - 1\right)}}{{\left(x - \sqrt{x^{2} + 1}\right)}^{2} + 1} + \frac{1}{x^{2} + 1}
=
0
.
$$
So the expression must be a constant. Taking $x=0$ and you will get $\pi/2$.
No need to think.
A: Colored for an easier reading. Let $\alpha= \arctan x $ and $\beta = \arctan(\sqrt{1 + x^2} - x)$ then $\alpha,\beta \in (-\pi/2,\pi/2)$ and we have $\tan \alpha = x$ and \begin{eqnarray}\tan \beta 
&=& \sqrt{1 + x^2} - x \\ 
&=& \color{red}{{1\over \cos \alpha}-\tan \alpha} \\
&=& {1-\sin\alpha \over  \cos\alpha} \\ 
&=&\color{red}{{(\cos (\alpha/2)-\sin(\alpha/2))^2\over \cos^2 (\alpha/2) \sin^2(\alpha/2) }}\\
&=&{\cos (\alpha/2)-\sin(\alpha/2)\over \cos (\alpha/2)+\sin(\alpha/2) }\\ &=&\color{red}{{1-\tan(\alpha/2)\over 1+\tan(\alpha/2) }}\\  
&=& \tan ({\pi\over 4}-\alpha/2) \end{eqnarray}
Since $\alpha,\beta \in (-\pi/2,\pi/2)$ we have $$\beta = {\pi\over 4}-\alpha/2$$ and we are done.
