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

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SUMMARY

The equation 2coth^(-1)(e^x) is proven to be equal to ln[cosech x + coth x]. The proof begins with the left-hand side (L.H.S.) and utilizes the identity 2coth^(-1)(e^x) = 2tanh^(-1)(e^-x). This is further simplified to ln[(1 + e^-x)/(1 - e^-x)], which is then shown to be equivalent to the right-hand side (R.H.S.) through substitution and manipulation. The hint to substitute u = e^x clarifies the relationship between the two sides.

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  • Understanding of hyperbolic functions, specifically coth and tanh.
  • Familiarity with inverse hyperbolic functions and their properties.
  • Knowledge of logarithmic identities and simplifications.
  • Ability to perform algebraic manipulations involving exponential functions.
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  • Study the properties of inverse hyperbolic functions, focusing on coth and tanh.
  • Learn about logarithmic identities and how to manipulate them in proofs.
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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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