Sum of Arcsecant and Arccosecant

Theorem
Let $x \in \R$ be a real number such that $\left|{x}\right| \ge 1$.

Then:
 * $\operatorname{arcsec} x + \operatorname{arccsc} x = \dfrac \pi 2$

where $\operatorname{arcsec}$ and $\operatorname{arccsc}$ denote arcsecant and arccosecant respectively.

Proof
Let $y \in \R$ such that:
 * $\exists x \in \R, \left|{x}\right| \ge 1: x = \csc \left({y + \dfrac \pi 2}\right)$

Then:

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

Then we can write $-y = \operatorname{arccsc} x$.

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

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

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

That is, $\dfrac \pi 2 = \operatorname{arcsec} x + \operatorname{arccosec} x$.