Prove 2coth^(-1)(e^x) = ln[cosech x + coth x]

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To prove the equation 2coth^(-1)(e^x) = ln[cosech x + coth x], the initial approach involves manipulating the left-hand side (LHS) by expressing it in terms of tanh^(-1). The transformation leads to ln[(1+(e^-x))/(1-(e^-x))], but progress stalls at this point. A suggestion is made to work from the right-hand side (RHS) and use the substitution u = e^x to simplify the process. This substitution is expected to clarify the relationship between the two sides of the equation. The discussion emphasizes the importance of strategic manipulation in proving the identity.
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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.
 
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sharks said:
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.

Hints: work from the RHS, and substitute u = e^x. Will make everything much clearer.
 

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