$\arctan (x) + \arctan(1/x) = \frac{\pi}{2}$ 
How can I show that $\arctan (x) + \arctan(1/x) =\frac{\pi}{2}$?

I tried to let $x = \tan(u)$. Then 
$$ \arctan(\tan(u)) + \arctan(\tan(\frac{\pi}{2} - x)) = \frac{\pi}{2}$$
but it does not seem useful. 

I'd appreciate most a proof that gives intuition and / or uses geometric
  insight.


$\arctan(a)+\arctan(b)
=\arctan(\frac{a+b}{1-ab})
$.
Therefore
$\arctan(x)+\arctan(1/x)
=\arctan(\frac{x+1/x}{1-1})
=\arctan(\frac{x+1/x}{0})
=\pi/2
$.
If this bothers you,
$\begin{array}\\
\arctan(x-c)+\arctan(1/x)
&=\arctan(\frac{x-c+1/x}{1-(x-c)/x)})\\
&=\arctan(\frac{x-c+1/x}{c/x})\\
&=\arctan(\frac{x^2-cx+1}{c})\\
&=\arctan(\frac{x^2+1}{c}-x)\\
&\to \pi/2
\quad\text{ as } c \to 0\\
\end{array}
$.
