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Derivative of ArcCosh[x/a] for x>a>0 is 1/Sqrt[x^2-a^2]. Derivative of ArcTanh[x/Sqrt[x^2-a^2]] is also 1/Sqrt[x^2-a^2]. However, I am not able to prove that ArcCosh[x/a]=ArcTanh[x/Sqrt[x^2-a^2]]+constant Any help??

sumant ola
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    You have proven that they are equal up to a constant (assuming it's correct, I haven't checked). Maybe you meant to say that you want to prove it through trigonometry rather than calculus? – Arthur Oct 25 '23 at 06:48
  • Related: https://math.stackexchange.com/q/2183650/42969 – Martin R Oct 25 '23 at 06:51
  • Arthur, this is what Mathematica gives. I would like to prove it using trigonometry. – sumant ola Oct 25 '23 at 08:44

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