Shape of Cosecant Function

Theorem
The cosecant function is:


 * $(1): \quad$ strictly decreasing on the intervals $\left[{- \dfrac \pi 2 \,.\,.\, 0}\right)$ and $\left({0 \,.\,.\, \dfrac \pi 2}\right]$
 * $(2): \quad$ strictly increasing on the intervals $\left[{\dfrac \pi 2 \,.\,.\, \pi}\right)$ and $\left({\pi \,.\,.\, \dfrac {3 \pi} 2}\right]$

Proof
From Derivative of Cosecant Function: $D_x \left({\csc x}\right) = -\dfrac{\cos x}{\sin^2 x}$

From Sine and Cosine are Periodic on Reals/Corollary: $\forall x \in \left({- \dfrac \pi 2 \,.\,.\, \dfrac {3 \pi} 2}\right) \setminus \left\{0, \pi\right\}: \sin x \ne 0$.

Thus, from Square of Element of Ordered Integral Domain is Positive: $\forall x \in \left({- \dfrac \pi 2 \,.\,.\, \dfrac {3 \pi} 2}\right) \setminus \left\{0, \pi\right\}: \sin^2 x > 0$.

From Sine and Cosine are Periodic on Reals/Corollary: $\cos x > 0$ on the open interval $\left({- \dfrac \pi 2 \,.\,.\, \dfrac \pi 2}\right)$.

It follows that $\forall x \in \left({- \dfrac \pi 2 \,.\,.\, \dfrac \pi 2}\right) \setminus \left\{0\right\}: -\dfrac{\cos x}{\sin^2 x} < 0$.

From Sine and Cosine are Periodic on Reals/Corollary: $\cos x < 0$ on the open interval $\left({\dfrac \pi 2 \,.\,.\, \dfrac {3 \pi} 2}\right)$.

It follows that $\forall x \in \left({\dfrac \pi 2 \,.\,.\, \dfrac {3 \pi} 2}\right) \setminus \left\{\pi\right\}: -\dfrac{\cos x}{\sin^2 x} > 0$.

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

Also see

 * Shape of Sine Function
 * Shape of Tangent Function
 * Shape of Cotangent Function