# Shape of Secant Function

## Theorem

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

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

## Proof

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

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

It follows that:

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

It follows that:

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

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

$\cos - \dfrac \pi 2 = \cos \dfrac \pi 2 = \cos \dfrac {3 \pi} 2 = 0$
$\cos x > 0$ on the open interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$

From the same source:

$\cos x < 0$ on the open interval $\openint {\dfrac \pi 2} {\dfrac {3 \pi} 2}$

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

### Graph of Secant Function

$\blacksquare$