Shape of Cosecant Function

Theorem

The nature of the cosecant function on the set of real numbers $\R$ is as follows:

$(1): \quad$ strictly decreasing on the intervals $\hointr {-\dfrac \pi 2} 0$ and $\hointl 0 {\dfrac \pi 2}$
$(2): \quad$ strictly increasing on the intervals $\hointr {\dfrac \pi 2} \pi$ and $\hointl \pi {\dfrac {3 \pi} 2}$
$(3): \quad$ $\csc x \to +\infty$ as $x \to 0^+$
$(4): \quad$ $\csc x \to +\infty$ as $x \to \pi^-$
$(5): \quad$ $\csc x \to -\infty$ as $x \to \pi^+$
$(6): \quad$ $\csc x \to -\infty$ as $x \to 2 \pi^-$

Proof

$\map {D_x} {\csc x} = -\dfrac {\cos x} {\sin^2 x}$
$\forall x \in \openint {-\dfrac \pi 2} {\dfrac {3 \pi} 2} \setminus \set {0, \pi}: \sin x \ne 0$
$\forall x \in \openint {-\dfrac \pi 2} {\dfrac {3 \pi} 2} \setminus \set {0, \pi}: \sin^2 x > 0$

$\cos x > 0$ on the open interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$

It follows that:

$\forall x \in \openint {-\dfrac \pi 2} {\dfrac \pi 2} \setminus \set 0: -\dfrac {\cos x} {\sin^2 x} < 0$
$\cos x < 0$ on the open interval $\openint {\dfrac \pi 2} {\dfrac {3 \pi} 2}$

It follows that:

$\forall x \in \openint {\dfrac \pi 2} {\dfrac {3 \pi} 2} \setminus \set \pi: -\dfrac {\cos x} {\sin^2 x} > 0$

Thus, $(1)$ and $(2)$ follow from Derivative of Monotone Function.

From Zeroes of Sine and Cosine: $\sin 0 = \sin \pi = \sin 2 \pi = 0$.

$\sin x > 0$ on the open interval $\openint 0 \pi$

From the same source:

$\sin x < 0$ on the open interval $\openint \pi {2 \pi}$

Thus, $(3)$, $(4)$, $(5)$ and $(6)$ follow from Infinite Limit Theorem.

Graph of Cosecant Function

$\blacksquare$