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[alyafey22]

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$$\sinh \left( \tanh^{-1} (x) \right) = \frac{x}{ \sqrt{1-x^2}}$$

 

[alyafey22]

Hey Mohammed , We know that :$$\sinh(x) = \frac{e^{x}-e^{-x}}{2}$$

$$\sinh\left( \tanh^{-1} (x) \right) = \frac{e^{\tanh^{-1} (x)}-e^{-\tanh^{-1} (x)}}{2}$$

We know also :

$$\tanh^{-1} (x) = \ln \left( \sqrt{ \frac{1+x}{1-x}} \right)$$

$$\sinh \left( \tanh^{-1} (x) \right) = \frac{ \sqrt{ \frac{1+x}{1-x} }- \sqrt{ \frac{1-x}{1+x} } }{2}$$

$$\sinh \left( \tanh^{-1} (x) \right) = \frac{x}{ \sqrt{1-x^2}}$$

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