I tried to get the trig function cosec in MathJax, but it comes out as $\cosec$. In contrast, it works perfectly well for sec: $\sec$. Of course, one can always use \operatorname{cosec} as a workaround ($\operatorname{cosec})$; but this seems clumsy. The shorter form csc works: $\csc$. Perhaps the designers of MathJax simply don't like cosec and think that we should all switch to csc.

  • $\begingroup$ This is not a bug wrt Stack Exchange software, it is a feature of mathjax, over which Stack Exchange has no control. Mathjax is not unique to some stack exchange sites. $\endgroup$
    – amWhy
    Apr 2, 2020 at 17:05
  • 1
    $\begingroup$ It also recognizes $\sinh x$, but not $\sech x$. $\endgroup$ Apr 2, 2020 at 21:44
  • 1
    $\begingroup$ Related (a bit): Request for Hyperbolic $ \sech \,x $ inclusion in MathJax. $\endgroup$ Apr 3, 2020 at 4:51

1 Answer 1


No, you're right that mathjax doesn't recognize $\cosec(x)$: $\cosec(x)$, but it recognizes $\csc(x)$: $\csc(x)$ for use when you want to denote cosecant(x). It is likely due to the fact that \sin(x), \cos(x), \tan(x), \sec(x), \cot(x) all have three-letter operator-names, and so the choice of operator for cosecant, in mathjax, similarly has a three-letter operator: \csc(x).

Even though one can speak of $\csc(x) = \frac 1{\sin x}$, mathjax does not recognize $\arccsc(x)$: $\arccsc(x)$, though it recognizes $\csc^{-1}(x)$: $\csc^{-1}(x)$. But nor does it recognize $\arcsec$: $\arcsec$, nor $\arccot$: $\arccot$.

But we do have $\arcsin, \arccos, \arctan$: $\arcsin, \arccos, \arctan$.

Now, a work around, if you're needing to use cosecant a lot in one post, is to write \DeclareMathOperator{cosec}{cosec} before needing to use cosec. Then, you can use it as you like, as you expected you could, by writing \cosec(x). I display the results immediately below. You can click on "edit" to see that it does indeed work.




You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .