The Inverse Secant $y = \operatorname{arcsec} x$

$\operatorname{arcsec} x$, the inverse secant, inverts $\sec x = \dfrac{1}{\cos x}$¹. Since $\sec$ is periodic, a range must be chosen before an inverse exists, and the usual choice is $[0, \pi]$, matching $\arccos$.

Definition and closed form

The equation $\sec y = x$ is the same as $\cos y = \dfrac{1}{x}$, so the function can be written in closed form.

Because $\arccos$ is defined on $[-1, 1]$, we need $\left|\dfrac{1}{x}\right| \leq 1$, that is $|x| \geq 1$.

Domain and range

• The domain is $x \leq -1$ or $x \geq 1$
• The range is $[0, \pi]$ with $\dfrac{\pi}{2}$ removed
• At $x = 1$ the value is $0$, and at $x = -1$ it is $\pi$

There is no curve on $-1 < x < 1$, which mirrors the fact that $|\sec y| \geq 1$ always. Attaining $\dfrac{\pi}{2}$ would require $\dfrac{1}{x} = 0$, so that value is out of reach too.

Asymptote

As $x \to +\infty$ and as $x \to -\infty$ alike, $\dfrac{1}{x} \to 0$ and hence $y \to \dfrac{\pi}{2}$, so the line $y = \dfrac{\pi}{2}$ is a horizontal asymptote. The right branch approaches it from below and the left branch from above, so the two branches sandwich it. Since the value $\dfrac{\pi}{2}$ is never attained, the asymptote is a boundary in the literal sense.

Notable values

Monotonicity and tangents

The derivative is as follows.

It is positive throughout the domain, so the function increases on both branches. As $|x| \to 1$ the factor $\sqrt{x^{2}-1}$ tends to $0$ and the derivative diverges, so the tangents at the endpoints $(1, 0)$ and $(-1, \pi)$ are vertical. The curve stands up at its ends and lies down along the asymptote far away.

No discontinuity

Because the domain excludes $-1 < x < 1$, this function has neither jumps nor oscillation. Each branch is continuous and their ranges do not overlap. It too is an inverse taken through a reciprocal, but it avoids the trouble that arises for $\operatorname{arccot}$, whose domain straddles the origin and where the choice of principal value becomes a genuine question.

Relation to the inverse cosecant

This is nothing but $\arccos u + \arcsin u = \dfrac{\pi}{2}$ with $u = \dfrac{1}{x}$. The two graphs are therefore reflections of one another in the line $y = \dfrac{\pi}{4}$.

1. Inverse trigonometric functions, Wikipedia