Prove that sec^-1(x) is equal to cos^-1(1/x)

The Mathematician — Sun, 11 Dec 2022 13:00:49 +0000

We will prove that

\sec^{-1}(x) = \cos^{-1}(1/x)

for

\lvert x \rvert \geq 1

.

Proof. We need to take two things into account:

\begin{align*} \cos^{-1}(x), \quad -1 \leq x \leq 1, \end{align*}

and

\begin{align*} \sec^{-1}(x), \quad \lvert x \rvert \geq 1. \end{align*}

Let

y = \sec^{-1}(x)

. Then

\begin{align*} y = \sec^{-1}(x) &\iff \sec(y) = x \\ &\iff \frac{1}{\cos(y)} = x \quad \text{since } \sec(y) = \frac{1}{\cos(y)} \\ &\iff \cos(y) = \frac{1}{x} \\ &\iff y = \cos^{-1}(1/x). \end{align*}

Since

y = \sec^{-1}(x)

, we have that

\sec^{-1}(x) = \cos^{-1}(1/x)

for

-1 \leq 1/x \leq 1

, which translates to

\lvert x \rvert \geq 1

sec^-1(x) = cos^-1(1/x) Archives - Epsilonify