Home

To verify the formula, we can use the identities:

$$\cos(2\theta)=\frac{1+\tan^2\theta}{1+\tan^2\theta}=\frac{1-\tanh^2\theta}{1+\tanh^2\theta}$$

and

$$\sin(2\theta)=\frac{2\tan\theta}{1+\tan^2\theta}=\frac{2\tanh\theta}{1+\tanh^2\theta}.$$

Using these identities, we have:

$$\begin{align*} \frac{1-\tanh^2\ln(1+x^2)}{1+\tanh^2\ln(1+x^2)} &= \frac{1-\tanh^2(u)}{1+\tanh^2(u)} &= \cos(2u) &= \cos\left(2\arctan\left(\frac{x}{\sqrt{1+x^2}}\right)\right) &= \frac{1-x^2}{1+x^2}. \end{align*}$$

Similarly,

$$\begin{align*} \frac{2\tanh\ln(1+x^2)}{1+\tanh^2\ln(1+x^2)} &= \frac{2\tanh(u)}{1+\tanh^2(u)} &= \sinh(2u) &= \sin\left(2\arctan\left(\frac{x}{\sqrt{1+x^2}}\right)\right) &= \frac{2x}{1+x^2}. \end{align*}$$

Therefore, we have:

$$\tanh(\ln(1+x^2))=\frac{1-x^2}{1+x^2}+\frac{2x}{1+x^2}i.$$