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Why (correct) expressions like

Assuming[p > 0, 2 ArcTan[Sinh[p]] == Pi - 2 ArcTan[Csch[p]] // FullSimplify]

Are not correctly evaluated to: True? What is the best approach in these cases

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    $\begingroup$ Note that the simpler expression Assuming[p > 0, 2 ArcTan[p] == Pi - 2 ArcTan[1/p] // FullSimplify] does evaluate to True. The problem seems to be that Mathematica doesn't recognize $\sinh x = 1/\mathrm{csch}\, x$ in this context. $\endgroup$ Mar 14, 2019 at 13:57
  • $\begingroup$ Series shows that the difference between the LHS and RHS is zero, i.e., Assuming[p > 0, Series[ 2 ArcTan[Sinh[p]] - (Pi - 2 ArcTan[Csch[p]]), {p, 0, 50}]] // Normal evaluates to 0 $\endgroup$
    – Bob Hanlon
    Nov 14, 2019 at 1:51

1 Answer 1

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Your have to help a bit to FoolSimplify[]

f[e_] := 100 Count[e, _Gudermannian | _Csch, {0, Infinity}] + 
      LeafCount[e]

Assuming[p > 0, 
 FullSimplify[2 ArcTan[Sinh[p]] == Pi - 2 ArcTan[Csch[p]], 
  ComplexityFunction -> f]]

(True)

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