:::section[Two Different Composition Questions] For an inverse pair $f$ and $f^{-1}$, the expression $f^{-1}(f(x))$ is governed by the chosen branch of $f$, whereas $f(f^{-1}(y))$ is governed by the domain of $f^{-1}$. The restrictions must be stated separately. ::: :::theorem[Composition Laws for Sine, Cosine, and Tangent] $$\begin{align*} \sinh^{-1}(\sinh x)&=x, &&x\in\mathbb{R},\\ \sinh(\sinh^{-1}x)&=x, &&x\in\mathbb{R},\\ \cosh^{-1}(\cosh x)&=x, &&x\in[0,\infty),\\ \cosh(\cosh^{-1}x)&=x, &&x\in[1,\infty),\\ \tanh^{-1}(\tanh x)&=x, &&x\in\mathbb{R},\\ \tanh(\tanh^{-1}x)&=x, &&x\in(-1,1). \end{align*}$$ ::: :::proof[Why the Cosine Restriction Is Necessary] The principal inverse $\cosh^{-1}$ has range $[0,\infty)$. Since $\cosh$ is even, $$\begin{equation*} \cosh^{-1}(\cosh x)=|x| \end{equation*}$$ for arbitrary real $x$. Therefore it equals $x$ precisely on the selected branch $x\ge0$. In the reverse composition, every $x\ge1$ lies in the domain of $\cosh^{-1}$, so $\cosh(\cosh^{-1}x)=x$. $\square$ ::: :::theorem[Composition Laws for Cotangent, Secant, and Cosecant] With the principal real branches, $$\begin{align*} \coth^{-1}(\coth x)&=x, &&x\in\mathbb{R}\setminus\{0\},\\ \coth(\coth^{-1}x)&=x, &&|x|>1,\\ \operatorname{sech}^{-1}(\operatorname{sech}x)&=x, &&x\in[0,\infty),\\ \operatorname{sech}(\operatorname{sech}^{-1}x)&=x, &&0<x\le1,\\ \operatorname{csch}^{-1}(\operatorname{csch}x)&=x, &&x\ne0,\\ \operatorname{csch}(\operatorname{csch}^{-1}x)&=x, &&x\ne0. \end{align*}$$ ::: :::example[Composition Outside the Principal Branch] Take $x=-2$. Then $$\begin{equation*} \cosh^{-1}(\cosh(-2))=2\ne-2. \end{equation*}$$ The output must lie in the principal range $[0,\infty)$. ::: :::mistake[Do Not Cancel Functions Without Checking Domains] Writing $\cosh^{-1}(\cosh x)=x$ for every real $x$ ignores the fact that $\cosh$ is not one-to-one on $\mathbb{R}$. “Cancellation” is valid only on the branch used to define the inverse. ::: :::exercise[Practice Questions] 1. Simplify $\cosh^{-1}(\cosh(-3))$. 2. State the domain on which $\operatorname{sech}(\operatorname{sech}^{-1}x)=x$. 3. Determine whether $\tanh^{-1}(\tanh 5)=5$. 4. Simplify $\operatorname{sech}^{-1}(\operatorname{sech}(-a))$ for real $a$. ::: :::answer[Answers and Guidance] 1. The value is $3$, because the principal inverse cosine has nonnegative range. 2. The domain is $(0,1]$. 3. Yes. The function $\tanh$ is one-to-one on all of $\mathbb{R}$. 4. Since $\operatorname{sech}$ is even and the inverse has range $[0,\infty)$, the value is $|a|$. ::: :::faq[Frequently Asked Questions] Q: Why do the two composition orders have different domains? A: In $f^{-1}(f(x))$, $x$ must lie in the branch domain of $f$. In $f(f^{-1}(x))$, $x$ must lie in the domain of the inverse, which is the range of that branch. Q: Is $\cosh^{-1}(\cosh x)$ always $|x|$? A: For the principal real branch, yes. Q: Why is zero excluded for inverse hyperbolic cosecant? A: The function $\operatorname{csch}x$ never takes the value zero, so zero cannot belong to the domain of its inverse. ::: :::call-to-action[Continue Learning] subtitle: Differentiate the inverse functions and track every branch, endpoint, and absolute-value condition. button: Differentiate Inverse Functions|/real-analysis/functions/hyperbolic-and-inverse-hyperbolic-functions/differentiation-of-inverse-hyperbolic-functions button-ghost: Review This Section|/real-analysis/functions/hyperbolic-and-inverse-hyperbolic-functions :::