Sum of Arcsecant and Arccosecant

Theorem

Let $x \in \R$ be a real number such that $\size x \ge 1$.

Then:

$\arcsec x + \arccsc x = \dfrac \pi 2$

where $\arcsec$ and $\arccsc$ denote arcsecant and arccosecant respectively.

Let $y \in \R$ such that:

$\exists x \in \R: \size x \ge 1$ and $x = \map \csc {y + \dfrac \pi 2}$

Then:

$\ds x$

$=$

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

$\ds $

$=$

$\ds -\csc y$

$\ds $

$=$

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

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

Then we can write:

$-y = \arccsc x$

But then:

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

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

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

Hence:

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

That is:

$\dfrac \pi 2 = \arcsec x + \arccsc x$

$\blacksquare$