Sum of Arctangent and Arccotangent

Theorem

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

Then:

$\arctan x + \arccot x = \dfrac \pi 2$

where $\arctan$ and $\arccot$ denote arctangent and arccotangent respectively.

Let $y \in \R$ such that:

$\exists x \in \R: x = \map \cot {y + \dfrac \pi 2}$

Then:

$\ds x$

$=$

$\ds \map \cot {y + \frac \pi 2}$

$\ds $

$=$

$\ds -\tan y$

$\ds $

$=$

$\ds \map \tan {-y}$

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

Then we can write:

$-y = \arctan x$

But then:

$\map \cot {y + \dfrac \pi 2} = x$

Now because $-\dfrac \pi 2 \le y \le \dfrac \pi 2$ it follows that:

$0 \le y + \dfrac \pi 2 \le \pi$

Hence:

$y + \dfrac \pi 2 = \arccot x$

That is:

$\dfrac \pi 2 = \arccot x + \arctan x$

$\blacksquare$