<s>[INST] $\arctan (x) + \arctan(1/x) = \frac{\pi}{2}$ 
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.

 [/INST] Let $f(x)=\arctan(x)+\arctan(\frac1x)$. Then $f'(x)=\frac{1}{1+x^2}+\frac{1}{1+1/x^2}\frac{-1}{x^2}=\frac{1}{1+x^2}-\frac{1}{1+x^2}=0$ so $f(x)$ is constant for $x>0$. Then note that $f(1)=\frac{\pi}{2}$ and thus $f(x)=\frac{\pi}{2}$.
There is a discontinuity at $x=0$, so the derivative only makes sense for $x\neq 0$. When $x<0$, you can check that $f(-1)=-\frac{\pi}{2}$. $f(x)$ has the same derivative so for $x<0$, $f(x) -\frac{\pi}{2}$
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