The Secant Function

Definition For a real number $x$, measured in radians, with $\cos x \ne 0$, the secant of $x$ is

Secant has the same sign as cosine, positive where the unit circle point lies to the right of the vertical axis and negative where it lies to the left.

Definition Secant is undefined exactly where $\cos x = 0$, which is at $x = \frac{\pi}{2} + k\pi$ for integers $k$. Its domain is

Definition The range of secant is

For every $x$ in the domain, $0 < \lvert \cos x \rvert \le 1$, so $\lvert \sec x \rvert = \frac{1}{\lvert \cos x \rvert} \ge 1$. Conversely, if $\lvert y \rvert \ge 1$ then $\frac{1}{y}$ is a nonzero number in $[-1, 1]$, so some $x$ has $\cos x = \frac{1}{y}$ and then $\sec x = y$. Secant has no zeros, because a fraction with numerator $1$ is never $0$.

Definition The line $x = a$ is a vertical asymptote of the graph of $f$ if $f(x)$ grows without bound, toward $\infty$ or $-\infty$, as $x$ approaches $a$ from at least one side. Secant has a vertical asymptote at every excluded input $x = \frac{\pi}{2} + k\pi$, because there $\cos x$ approaches $0$ and its reciprocal grows without bound. Cosine changes sign at each of these inputs, so secant goes to $\infty$ on one side and to $-\infty$ on the other. The branches on the two sides are separate pieces of the graph and must not be joined across the asymptote.

Definition For every $x$ in the domain, $-x$ is also in the domain and

So secant is an even function, and its graph is symmetric about the vertical axis.

Definition For every $x$ in the domain, $x + 2\pi$ is in the domain and $\sec(x + 2\pi) = \sec x$, because $\cos(x + 2\pi) = \cos x$. The least positive period is $2\pi$. If some $p$ with $0 < p < 2\pi$ were a period, then $p = 0 + p$ would be in the domain because $0$ is, and $\sec p = \sec 0 = 1$, which needs $\cos p = 1$. No number strictly between $0$ and $2\pi$ has cosine $1$, so no such $p$ exists. Adding only $\pi$ flips the sign, $\sec(x + \pi) = -\sec x$.

Definition The graph of $y = \sec x$ is made of separate branches, one on each open interval between consecutive asymptotes. On the central branch $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ the cosine satisfies $0 < \cos x \le 1$, with $\cos x = 1$ only at $x = 0$. Taking reciprocals reverses the inequality between positive numbers, so

with equality only at $x = 0$. The central branch is a U shape with lowest point $(0, 1)$. On the next branch $\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)$ the cosine satisfies $-1 \le \cos x < 0$, so $\sec x \le -1$, with highest point $(\pi, -1)$. The branches then alternate between opening upward and opening downward every $\pi$. No derivative is needed for these extreme points, only the bounds on cosine.