Question

Tanh(lnx) = x^2-1/x^2+1 for x > 0

Answer 1

\[\tanh(lnx) = \frac{ x^2-1 }{ x^2-1 } for x > 0\]

Answer 2

Prove the following identity

Answer 3

I don't know how , batman >.<

Answer 4

I'd just start with the definition. You can simplify all of that \(e^{\ln x}\) crap pretty easily and then it's just algebra.