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?

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.
