Sum of Arctangent and Arccotangent

Theorem
Let $x \in \R$ be a real number.

Then:
 * $\arctan x + \operatorname{arccot} x = \dfrac \pi 2$

where $\arctan$ and $\operatorname{arccot}$ denote arctangent and arccotangent respectively.

Proof
Let $y \in \R$ such that:
 * $\exists x \in \R: x = \cot \left({y + \dfrac \pi 2}\right)$

Then:

Suppose $-\dfrac \pi 2 \le y \le \dfrac \pi 2$.

Then we can write $-y = \arctan x$.

But then $\cot \left({y + \dfrac \pi 2}\right) = x$.

Now since $-\dfrac \pi 2 \le y \le \dfrac \pi 2$ it follows that $0 \le y + \dfrac \pi 2 \le \pi$.

Hence $y + \dfrac \pi 2 = \operatorname{arccot} x$.

That is, $\dfrac \pi 2 = \operatorname{arccot} x + \arctan x$.