# Let $f(x) = \arctan x + \mathop{\text{arccot}} x$. Find $f(0) + f(1) + f(\sqrt{2}) + f(\sqrt{3})$.

Let $f(x) = \arctan x + \mathop{\text{arccot}} x$. Find $f(0) + f(1) + f(\sqrt{2}) + f(\sqrt{3})$.

This is what i got,

$$\left(0+\frac{\pi}{2}\right)+\left(\frac{\pi}{4}+\frac{\pi}{4}\right)$$

• I take it those are the first two terms $f(0)$ and $f(1)$. Hmm ... something interesting there: they both equal $\pi/2$. I wonder if $f(x)$ for arbitrary $x$ follows such a pattern.... – Simon S Jun 5 '15 at 22:46
• You can construct an angle of given tangent x with a right triangle of legs x and 1. – Piquito Jun 5 '15 at 22:49
• should not each one of them be $\pi/2,$ so totals $2\pi$ – abel Jun 5 '15 at 22:56

We know that $\tan \theta = x$, so $\theta = \arctan x$. What is the other angle, as a function of (a) $\theta$ and (b) $x$ ?
$\arctan x+ \operatorname{arccot} x =\frac\pi2$