DryRun
Gold Member
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Homework Statement
Prove 2coth^(-1)(e^x) = ln[cosech x + coth x]
The attempt at a solution
I'm starting with the L.H.S. first, which is conventional, i think.
2coth^(-1)(e^x)
= 2tanh^(-1)(e^-x)
= 2[1/2 ln [(1+(e^-x))/(1-(e^-x))]
= ln [(1+(e^-x))/(1-(e^-x))]
I'm stuck.
So, i tried to cheat a little, by expanding the R.H.S. and see if i could find a way to convert the last step above into the R.H.S. but my copybook has become something of a painting.
Prove 2coth^(-1)(e^x) = ln[cosech x + coth x]
The attempt at a solution
I'm starting with the L.H.S. first, which is conventional, i think.
2coth^(-1)(e^x)
= 2tanh^(-1)(e^-x)
= 2[1/2 ln [(1+(e^-x))/(1-(e^-x))]
= ln [(1+(e^-x))/(1-(e^-x))]
I'm stuck.
So, i tried to cheat a little, by expanding the R.H.S. and see if i could find a way to convert the last step above into the R.H.S. but my copybook has become something of a painting.