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

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In summary, the conversation discusses proving the identity 2coth^(-1)(e^x) = ln[cosech x + coth x]. The attempt at a solution involves manipulating the LHS using the inverse hyperbolic tangent function and getting stuck. The suggestion is made to work from the RHS and substitute u = e^x to simplify the problem.
  • #1
DryRun
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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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  • #2
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.
 

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

What is the meaning of "2coth^(-1)(e^x) = ln[cosech x + coth x]"?

The equation is a mathematical expression that relates the inverse hyperbolic cotangent function to the natural logarithm of the sum of the hyperbolic cosecant and cotangent functions.

How is the equation derived?

The equation can be derived using the properties of hyperbolic functions and the definition of inverse functions. It involves using substitution and simplification techniques to reach the final result.

What are the applications of this equation?

This equation has applications in mathematical analysis, physics, and engineering. It can be used to solve problems involving hyperbolic functions, such as finding the derivative or integral of a function containing these functions.

Is the equation valid for all values of x?

No, the equation is only valid for values of x that make both sides of the equation defined. This means that x cannot be equal to zero, and the argument of the logarithmic and inverse hyperbolic functions must be positive.

Can the equation be simplified further?

Yes, the equation can be simplified by using algebraic manipulations and the properties of logarithms and hyperbolic functions. However, the simplified form may not be as useful or informative as the original equation.

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