If $\theta=\ln(\tan\alpha)$, then $\tanh\theta$ simplifies to:

1. Use $\tanh\theta=\dfrac{e^{\theta}-e^{-\theta}}{e^{\theta}+e^{-\theta}}$ with $e^{\theta}=\tan\alpha$. 2. That gives $\dfrac{\tan\alpha-\frac{1}{\tan\alpha}}{\tan\alpha+\frac{1}{\tan\alpha}}=\dfrac{\tan^2\alpha-1}{\tan^2\alpha+1}$. 3. Write $\tan^2\alpha=\dfrac{\sin^2\alpha}{\cos^2\alpha}$ and clear the fractions: $\dfrac{\sin^2\alpha-\cos^2\alpha}{\sin^2\alpha+\cos^2\alpha}$. 4. The denominator is 1 by the Pythagorean identity. 5. And $\sin^2\alpha-\cos^2\alpha=-\cos 2\alpha$, so $\tanh\theta=-\cos 2\alpha$. The sign is what the double-angle identity supplies. _Source: NECTA ACSEE 2022 Advanced Mathematics 142/1, Question 2: Hyperbolic Functions_