Without using \hspace{…pt} how to make nice alignment of separate equations?

Question

My code excluding preamble (don't judge please):

\subsection{Inverse Hyperbolic Trigonometric Identities}
\raggedright\setlength{\parindent}{45pt}
    \begin{enumerate}[align=left,widest=a,labelindent=16mm,leftmargin=!]
    \incrvariable\x
    \setcounter{enumi}{\x}
    \item%194
        $
        \begin{aligned}[t]
        \sinh(\cosh^{-1}{x})=\sqrt{x^2-1} \hspace{20pt} &\text{Domain: } [1,\infty) \hspace{26.5pt} &&\text{Range: } [0,\infty)
        \end{aligned}
        $
    \incrvariable\x
    \item%195
        $
        \begin{aligned}[t]
        \cosh(\sinh^{-1}{x})=\sqrt{x^2+1} \hspace{20pt} &\text{Domain: } (-\infty,\infty) \hspace{10pt} &&\text{Range: } [1,\infty)
        \end{aligned}
        $
    \incrvariable\x
    \item%196
        $
        \begin{aligned}[t]
        \sinh(\tanh^{-1}{x})=\frac{x}{\sqrt{1-x^2}} \hspace{16.4pt} &\text{Domain: } (-1,1) \hspace{22pt} &&\text{Range: } (-\infty,\infty)
        \end{aligned}
        $
    \incrvariable\x
    \item%197
        $
        \begin{aligned}[t]
        \cosh(\tanh^{-1}{x})=\frac{1}{\sqrt{1-x^2}} \hspace{15.25pt} &\text{Domain: } (-1,1) \hspace{21.9pt} &&\text{Range: } [1,\infty)
        \end{aligned}
        $
    \incrvariable\x
    \item%198
        $
        \begin{aligned}[t]
        \tanh(\sinh^{-1}{x})=\frac{x}{\sqrt{1+x^2}} \hspace{16.72pt} &\text{Domain: } (-\infty,\infty) \hspace{9.5pt} &&\text{Range: } (-1,1)
        \end{aligned}
        $
    \incrvariable\x
    \item%199
        $
        \begin{aligned}[t]
        \tanh(\cosh^{-1}{x})=\frac{\sqrt{x^2-1}}{x} \hspace{15.48pt} &\text{Domain: } [1,\infty) \hspace{26.1pt} &&\text{Range: } [0,1)
        \end{aligned}
        $
    \end{enumerate}

The \hspace{} is the important (and annoying) part.

How can I improve my code without changing "enumerate" and "aligned" or something else that looks almost identical? (Something that keeps track of elapsed pts?)

OR with preamble here: (it's looonnng)

\documentclass[12pt, twoside]{article}
\usepackage[utf8]{inputenc}
\usepackage{indentfirst}
\usepackage[left=1.2in,right=1.2in,top=1in,bottom=1in]{geometry}
\usepackage{titling}
\usepackage{titlesec}
\usepackage{pgfplots}
\usepackage{tikz}
\pgfplotsset{compat=1.18, width=10cm}
\usepackage[fleqn]{amsmath}
\usepackage{amssymb}
\usepackage{mathtools}
\usepackage{newunicodechar}
\DeclareMathOperator{\sgn}{sgn}
\newunicodechar{ℤ}{\mathbb{Z}}
\newunicodechar{ℝ}{\mathbb{R}}
\usepackage{enumitem}
\usepackage{graphicx}
\setlength\mathindent{3cm}
\usepackage{etoolbox}
\usepackage{microtype}
\usepackage{array}
\usepackage{booktabs}
\usepackage{xcolor}
\usepackage{romannum}
\usepackage{tocloft} %Change font size of TOC
%\renewcommand*{\arraystretch}{1.9}
\newcommand{\ra}[1]{\renewcommand{\arraystretch}{#1}}
\newcolumntype{C}{>{$\displaystyle}c<{$}}
\renewcommand{\contentsname}{Table of Content}
\DeclareMathOperator{\sech}{sech}
\DeclareMathOperator{\csch}{csch}
\usepgfplotslibrary{fillbetween}
\AtBeginDocument{\addtocontents{toc}{\small}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Odd page number on the right 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\def\align@preamble{%
   &\hfil
\strut@
\setboxz@h{@lign$\m@th\displaystyle{##}$}%
\ifmeasuring@\savefieldlength@\fi
\llap{\set@field}%
\tabskip\z@skip
   &\setboxz@h{@lign$\m@th\displaystyle{{}##}$}%
\ifmeasuring@\savefieldlength@\fi
\rlap{\set@field}
\hfil
\tabskip\alignsep@
}
\titleformat{\section} 
{\normalfont\huge\bfseries}{\makebox[45pt][l]{\thesection}}{0pt}{} 
\titleformat{\subsection} 
{\normalfont\large\bfseries}{\makebox[45pt][l]{\thesubsection}}{0pt}{}
\titleformat{\subsubsection} 
{\normalfont\normalsize\bfseries}{\makebox[45pt][l]{\thesubsubsection}}{0pt}{}
\setlength{\droptitle}{-1em}
\setlength{\parindent}{45pt}
\title{\Huge{\textbf{Mathematics Directory}}}
\author{\Large{Zhiyuan Liu}}
\date{}
%\newcommand\X{0}
%\renewcommand\X{0} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\def\newvariable#1{\gdef#1{0}}
\def\addtovariable#1#2{\xdef#1{\number\numexpr#1+#2\relax}}
\def\incrvariable#1{\addtovariable#1{1}}
\newvariable\x
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}
\maketitle
\section{}
\subsection{Inverse Hyperbolic Trigonometric Identities}
\raggedright\setlength{\parindent}{45pt}
    \begin{enumerate}[align=left,widest=a,labelindent=16mm,leftmargin=!]
    \incrvariable\x
    \setcounter{enumi}{\x}
    \item%194
        $
        \begin{aligned}[t]
        \sinh(\cosh^{-1}{x})=\sqrt{x^2-1} \hspace{20pt} &\text{Domain: } [1,\infty) \hspace{26.5pt} &&\text{Range: } [0,\infty)
        \end{aligned}
        $
    \incrvariable\x
    \item%195
        $
        \begin{aligned}[t]
        \cosh(\sinh^{-1}{x})=\sqrt{x^2+1} \hspace{20pt} &\text{Domain: } (-\infty,\infty) \hspace{10pt} &&\text{Range: } [1,\infty)
        \end{aligned}
        $
    \incrvariable\x
    \item%196
        $
        \begin{aligned}[t]
        \sinh(\tanh^{-1}{x})=\frac{x}{\sqrt{1-x^2}} \hspace{16.4pt} &\text{Domain: } (-1,1) \hspace{22pt} &&\text{Range: } (-\infty,\infty)
        \end{aligned}
        $
    \incrvariable\x
    \item%197
        $
        \begin{aligned}[t]
        \cosh(\tanh^{-1}{x})=\frac{1}{\sqrt{1-x^2}} \hspace{15.25pt} &\text{Domain: } (-1,1) \hspace{21.9pt} &&\text{Range: } [1,\infty)
        \end{aligned}
        $
    \incrvariable\x
    \item%198
        $
        \begin{aligned}[t]
        \tanh(\sinh^{-1}{x})=\frac{x}{\sqrt{1+x^2}} \hspace{16.72pt} &\text{Domain: } (-\infty,\infty) \hspace{9.5pt} &&\text{Range: } (-1,1)
        \end{aligned}
        $
    \incrvariable\x
    \item%199
        $
        \begin{aligned}[t]
        \tanh(\cosh^{-1}{x})=\frac{\sqrt{x^2-1}}{x} \hspace{15.48pt} &\text{Domain: } [1,\infty) \hspace{26.1pt} &&\text{Range: } [0,1)
        \end{aligned}
        $
    \end{enumerate}
\end{document}

OR someone may have already answered it. Tell me the post. Cheers.

Answer 1

I can think of two ways to accomplish your typesetting objective: with an alignat* environment, and with a longtable environment.

First, the alignat*-based approach. Observe: not a single \hspace statement anywhere in sight.

\documentclass[12pt, twoside]{article}
\usepackage[hmargin=1.2in,vmargin=1in]{geometry}
\usepackage{amsmath}
\allowdisplaybreaks
\newcounter{mycount}
\newcommand{\increment}{\refstepcounter{mycount}\themycount.}
\begin{document}
\setcounter{section}{4} % just for this example
\stepcounter{subsection}
\subsection{Inverse hyperbolic trigonometric identities}
\begin{alignat}{4}
&&&&& \text{Domain} && \text{Range} \[\jot] % "header line"
\increment &\qquad&
\sinh(\cosh^{-1}{x}) &= \sqrt{x^2-1} &\qquad&
[1,\infty) && [0,\infty) \
\increment &&
\cosh(\sinh^{-1}{x}) &= \sqrt{x^2+1} &&
(-\infty,\infty) &\qquad&
[1,\infty) \ \increment &&
\sinh(\tanh^{-1}{x}) &= \frac{x}{\sqrt{1-x^2}} &&
(-1,1) && (-\infty,\infty) \[\jot] % "\jot" inserts a bit more vertical whitespace
\increment &&
\cosh(\tanh^{-1}{x}) &= \frac{1}{\sqrt{1-x^2}} &&
(-1,1) && [1,\infty) \[\jot]
\increment &&
\tanh(\sinh^{-1}{x}) &= \frac{x}{\sqrt{1+x^2}} &&
(-\infty,\infty) && (-1,1) \[\jot]
\increment &&
\tanh(\cosh^{-1}{x}) &= \frac{\sqrt{x^2-1}}{x} &&
[1,\infty) && [0,1)
\end{alignat}
\end{document}

---

Second, the longtable-based approach. Observe: Still no \hspace directive anywhere in sight. :-)

\documentclass[12pt, twoside]{article}
\usepackage[hmargin=1.2in,vmargin=1in]{geometry}
\usepackage{amsmath}
\allowdisplaybreaks
\usepackage{longtable,array,booktabs}
\newcounter{mycount}
\newcommand{\increment}{\refstepcounter{mycount}\arabic{mycount}.}
\newcolumntype{L}{>{$\displaystyle}l<{$}}
\newcolumntype{Z}{>{\refstepcounter{mycount}\arabic{mycount}.}l}
\begin{document}
\setcounter{section}{4}   % just for this example
\stepcounter{subsection}
\subsection{Inverse hyperbolic trigonometric identities}
\begingroup
\renewcommand\arraystretch{1.5}
\addtolength\tabcolsep{5pt}
\begin{longtable}{@{} Z LLLL @{}}
%% header
\multicolumn{1}{@{}l}{} &&& \textup{Domain} & \textup{Range} \[1ex]
\endhead
%% body of table
& \sinh(\cosh^{-1}{x}) & \sqrt{x^2-1} &
  [1,\infty) & [0,\infty) \
& \cosh(\sinh^{-1}{x}) & \sqrt{x^2+1} &
  (-\infty,\infty) & [1,\infty) \ 
& \sinh(\tanh^{-1}{x}) & \frac{x}{\sqrt{1-x^2}} &
  (-1,1) & (-\infty,\infty) \ \addlinespace
& \cosh(\tanh^{-1}{x}) & \frac{1}{\sqrt{1-x^2}} & 
  (-1,1) & [1,\infty) \ \addlinespace
& \tanh(\sinh^{-1}{x}) & \frac{x}{\sqrt{1+x^2}} &
  (-\infty,\infty) & (-1,1) \ \addlinespace
& \tanh(\cosh^{-1}{x}) & \frac{\sqrt{x^2-1}}{x} &
  [1,\infty) & [0,1) \
\end{longtable}
\endgroup
\end{document}

Answer 2

You have a set of tabulated data. A simple tabular would do. I'd enclose each math expression between separate delimiters \(...\). If you have a large data to present that exceed a page or you simply expect a page break somewhere between lines, you would need longtable instead.

The example below is based on longtable. Uncomment \everymath={\displaystyle} if all math expressions should be in display style. This may require additional vertical spacing via increased arraystretch or extrarowheight.

\documentclass{article}
\usepackage{array}
\usepackage{amsmath}
\usepackage{longtable}
\begin{document}
\section{Long table}
\begingroup
\newcounter{eqno}\setcounter{eqno}{190}
\newcommand\addeqno{\stepcounter{eqno}\theeqno.}
\setlength\extrarowheight{6pt}
% \everymath={\displaystyle}
%%%
\begin{longtable}[r]{@{}
        r
        >{(}l<{)}
        >{Domain: (}l<{)}
        >{Range: (}l<{)}
    @{}}
    \addeqno & \sinh(\cosh^{-1}{x})=\sqrt{x^2-1}           & [1,\infty)       & [0,\infty) \
    \addeqno & \cosh(\sinh^{-1}{x})=\sqrt{x^2+1}           & (-\infty,\infty) & [1,\infty) \
    \addeqno & \sinh(\tanh^{-1}{x})=\frac{x}{\sqrt{1-x^2}} & (-1,1)           & (-\infty,\infty) \
    \addeqno & \cosh(\tanh^{-1}{x})=\frac{1}{\sqrt{1-x^2}} & (-1,1)           & [1,\infty) \
    \addeqno & \tanh(\sinh^{-1}{x})=\frac{x}{\sqrt{1+x^2}} & (-\infty,\infty) & (-1,1) \
    \addeqno & \tanh(\cosh^{-1}{x})=\frac{\sqrt{x^2-1}}{x} & [1,\infty)       & [0,1)
\end{longtable}
\endgroup
\end{document}

---

On the other hand, with enumerate, you would have to simulate columns. \makebox[<width>][<alignment>]{<content>} creates a box of a fixed width, no matter of the content. So if you place three such boxes horizontally, you achieve the desired effect. The downside is the width for each box has to be set manually and might need readjustment once you have changed the content. The package eqparbox provides a convenient macros, such as \eqmakebox[<label>][<alignment>}{<content>}. It works exactly as a regular makebox but instead of the width, user provides labels. Then, the width is calculated based on the longest content per label.

You can see this solution requires more coding, while the final effect is exactly the same as with longtable/tabular.

\documentclass{article}
\usepackage{enumitem}
\usepackage{eqparbox}
\usepackage{amsmath}
\begin{document}
\section{Enumerate}
\begin{enumerate}[
    align=left,
    widest=a,
    labelindent=10mm,
    leftmargin=!,
    before={\setcounter{enumi}{190}},
]
\item
    \eqmakebox[eq][l]{(\sinh(\cosh^{-1}{x})=\sqrt{x^2-1})}%
    \quad
    \eqmakebox[do][l]{Domain:([1,\infty))}%
    \quad
    \eqmakebox[ra][l]{Range:([0,\infty))}
\item
    \eqmakebox[eq][l]{(\cosh(\sinh^{-1}{x})=\sqrt{x^2+1})}%
    \quad
    \eqmakebox[do][l]{Domain: ((-\infty,\infty))}%
    \quad
    \eqmakebox[ra][l]{Range: ([1,\infty))}
\item
    \eqmakebox[eq][l]{(\sinh(\tanh^{-1}{x}) = \frac{x}{\sqrt{1-x^2}})}%
    \quad
    \eqmakebox[do][l]{Domain: ((-1,1))}%
    \quad
    \eqmakebox[ra][l]{Range: ((-\infty,\infty))}
\item
    \eqmakebox[eq][l]{(\cosh(\tanh^{-1}{x})=\frac{1}{\sqrt{1-x^2}})}%
    \quad
    \eqmakebox[do][l]{Domain: ((-1,1))}%
    \quad
    \eqmakebox[ra][l]{Range: ([1,\infty))}
\item
    \eqmakebox[eq][l]{(\tanh(\sinh^{-1}{x})=\frac{x}{\sqrt{1+x^2}})}%
    \quad
    \eqmakebox[do][l]{Domain: ((-\infty,\infty))}%
    \quad
    \eqmakebox[ra][l]{Range: ((-1,1))}
\item
    \eqmakebox[eq][l]{(\tanh(\cosh^{-1}{x})=\frac{\sqrt{x^2-1}}{x})}%
    \quad
    \eqmakebox[do][l]{Domain: ([1,\infty))}%
    \quad
    \eqmakebox[ra][l]{Range: ([0,1))}
\end{enumerate}
\end{document}