TeX - LaTeX Asked on December 11, 2020
I have the following table which for some reason doesn’t display the complete LHS vertical end line. I’ve notice that the line ‘broke’ when I added the ; I added
because I wanted some space between the end-horizontal line and the fractions. Could someone please help me to fix the line?
documentclass{article}
usepackage[utf8]{inputenc}
usepackage{amsmath}
usepackage{tabularx}
usepackage{graphicx}
begin{document}
begin{table}[h]
begin{center}
begin{tabular}{ | m{3.5cm} m{3.5cm} m{2.8cm}| }
hline
%vspace{2mm}
multicolumn{3}{|l|}{textbf{Reciprocal identities :}}
vspace{2mm} $sin theta = dfrac{1}{csc theta}$ & vspace{2mm} $cos theta = dfrac{1}{sec theta}$ & vspace{2mm} $tan theta = dfrac{1}{cot theta}$
vspace{3mm} $csc theta = dfrac{1}{sin theta}$ & vspace{2mm} $sec theta = dfrac{1}{cos theta}$ & vspace{2mm} $cot theta = dfrac{1}{tan theta}$
hline
end{tabular}
end{center}
end{table}
end{document}
Also, how can I add space between the horizontal line and ‘Reciprocal identities’? I tried vspace{2mm}
(with %
) but, once again, the line breaks and modifies the text to the center.
Thank you in advance.
You can use [Xpt]
instead of
at the end of the last table line for considering some space between the last table line and the horizontal line. In this way, you can also adjust the space by changing the X
value.
So, your code will be something like this:
documentclass{article}
usepackage[utf8]{inputenc}
usepackage{amsmath}
usepackage{tabularx}
usepackage{graphicx}
begin{document}
begin{table}[h]
begin{center}
begin{tabular}{ | m{3.5cm} m{3.5cm} m{2.8cm}| }
hline
%vspace{2mm}
multicolumn{3}{|l|}{textbf{Reciprocal identities :}}
vspace{2mm} $sin theta = dfrac{1}{csc theta}$ & vspace{2mm} $cos theta = dfrac{1}{sec theta}$ & vspace{2mm} $tan theta = dfrac{1}{cot theta}$
vspace{3mm} $csc theta = dfrac{1}{sin theta}$ & vspace{2mm} $sec theta = dfrac{1}{cos theta}$ & vspace{2mm} $cot theta = dfrac{1}{tan theta}$ [20pt]
hline
end{tabular}
end{center}
end{table}
end{document}
Correct answer by Venus on December 11, 2020
Here is my suggestion using framed
and an array
instead of the inner tabular:
documentclass{article}
usepackage[utf8]{inputenc}
usepackage{amsmath}
usepackage{framed}
begin{document}
begin{table}[h]
renewcommand{arraystretch}{2.5}
setlength{arraycolsep}{25pt}
begin{framed}
textbf{Reciprocal identities:}
[begin{array}{cll}
sin theta = dfrac{1}{csc theta}
& cos theta = dfrac{1}{sec theta}
& tan theta = dfrac{1}{cot theta}
csc theta = dfrac{1}{sin theta}
& sec theta = dfrac{1}{cos theta}
& cot theta = dfrac{1}{tan theta}
end{array}]
end{framed}
end{table}
end{document}
Answered by leandriis on December 11, 2020
You simply forgot, adding an empty row, to add a full empty row.
I propose two other solutions, one using cellspace
, which ensures a minimal spacing at the top and bottom of cells in columns with specifier prefixed with the letter S
(or C
if you load siunitx
and another entirely different, based on the empheq
package, which enables to have boxed multilined equations:
documentclass{article}
usepackage[utf8]{inputenc}
usepackage{empheq}
usepackage{tabularx}
usepackage{graphicx}
usepackage{cellspace}
setlength{cellspacetoplimit}{6pt}
setlength{cellspacebottomlimit}{6pt}
newcommand*widefbox[1]{setlength{fboxsep}{6pt}fbox{quad#1quad}}
begin{document}
begin{table}[h]
centering
begin{tabular}{ |*{2}{>{$}m{3.5cm}<{$}}>{$}m{2.8cm}<{$}| }
hline
multicolumn{3}{|l|}{textbf{Reciprocal identities :}}
vspace{2mm} sin θ= dfrac{1}{csc θ}&vspace{2mm} cos θ= dfrac{1}{sec θ} & vspace{2mm}tan θ= dfrac{1}{cot θ}
vspace{3mm} csc θ= dfrac{1}{sin θ} & vspace{2mm} sec θ= dfrac{1}{cos θ} & vspace{2mm} cot θ= dfrac{1}{tan θ} & &
hline
end{tabular}
vskip1cm
begin{tabular}{ |*{2}{>{$}S{m{3.5cm}}<{$}}>{$}S{m{2.8cm}}<{$}|}
hline
multicolumn{3}{|Sl|}{textbf{Reciprocal identities :}}
sin θ= dfrac{1}{csc θ}& cos θ= dfrac{1}{sec θ} &tan θ= dfrac{1}{cot θ}
csc θ= dfrac{1}{sin θ} & sec θ= dfrac{1}{cos θ} & cot θ= dfrac{1}{tan θ}
hline
end{tabular}
end{table}
begin{empheq}[box = widefbox]{align*}
& rlap{textbf{Reciprocal identities :}}
& sin θ= dfrac{1}{csc θ}& & cos θ= dfrac{1}{sec θ} & &tan θ= dfrac{1}{cot θ}[1.5ex]
& csc θ= dfrac{1}{sin θ} & & sec θ= dfrac{1}{cos θ} & & cot θ= dfrac{1}{tan θ}
end{empheq}
end{document}
Answered by Bernard on December 11, 2020
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