Another one from my assignment:
Assume $$ \tan (\frac{x}{2})=\tan A\tanh B $$
Prove that $$ \tan(x)=\frac{\sin(2A)\sinh(2B)}{1+cos(2A)\cosh(2B)} $$
I manipulated $$ \tan(x)=\tan2(\frac {x}{2})=\frac {2\tan A \tanh B}{1-(\tan A\tanh B)^2} $$ Using the half angle formula for tan. From there I seem to just be going in circles and coming back to the original question every attempt.
We will use the formulas in this page. Then $$\sin(2A)=2\sin(A)\cos(A)=\frac{2\tan(A)}{1/\cos^2(A)}=\frac{2\tan(A)}{1+\tan^2(A)}.$$ In a similar way we can express $\cos(2A)$, $\sinh(2B)$, $\cosh(2B)$ in terms of $\tan(A)$ or $\tanh(B)$.
Hence $$\tan(x)=\frac{\sin(2A)\sinh(2B)}{1+\cos(2A)\cosh(2B)}\\=\frac{2\tan A}{1+\tan^2 A}\cdot \frac{2\tanh B}{1-\tanh^2 B} \cdot \left(1+\frac{1-\tan^2 A}{1+\tan^2 A}\cdot\frac{1+\tanh^2 B}{1-\tanh^2 B}\right)^{-1}\\= \frac{2\tan A \tanh B}{1-(\tan A\tanh B)^2}.$$
On the other hand $$\tan(x)=\tan 2(x/2)=\frac{2\tan A \tanh B}{1-(\tan A\tanh B)^2}.$$