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# Inverse Hyperbolic Fuctions!! watch

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tanh^-1 ((x² - a²)/(x² + a²)) = ln(x/a)

help is greatly appreciated...will rep the person who gives correct solution.

(my rep is not worth much... but hey, its the thought that counts)
2. (Original post by Vishpatel)

tanh^-1 = ln(x/a)

help is greatly appreciated...will rep the person who gives correct solution.

(my rep is not worth much... but hey, its the thought that counts)
tanh(x)=e^x-e^(-x)/e^(x)+e^(-x)
tanh(ln(x/a)=[x/a-a/x]/[x/a+a/x]
=[x²-a²/ax]/[x²+a²/ax]
=(x² - a²)/(x² + a²)
3. tanh(ln(x/a)) = (e^[2ln(x/a)] - 1)/(e^[2ln(x/a)] + 1)
= [(x/a)^2 - 1]/[(x/a)^2 + 1]
= (x^2 - a^2)/(x^2 + a^2)
etc.

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Updated: January 25, 2006
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