How to prove sum related to hyperbolic tangents $\sum_{k=0}^{n-1}\frac{\tanh(...)}{1+\frac{\tanh^2x}{\tan^2(....)}}=\tanh(2nx)$

Question

I have no Idea how to start

I think to switch it to definite integral, use complex analysis, or some real analysis tricks and at the end I failed to make any progress.

$$ \displaystyle \sum_{k=0}^{n-1} \dfrac{\tanh \left( \dfrac x{n \sin^2 \left( \dfrac{2k+1}{4n} \pi\right)}\right)}{1 + \dfrac{\tanh^2 x}{\tan^2 \left( \dfrac{2k+1}{4n} \pi\right)}} = \tanh(2n x ) $$