Shape of Secant Function

Theorem
The secant function is:


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

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

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

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

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

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

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

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

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

Also see

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