trigonometry - How prove this trigonometric identity

Show that

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

Thank you ,This problem I take some hours,and at last I don't prove it

and This problem is from

This book have some same problem.all is not true? if not true,and we how find it or edit it somewhere?

Thank you achille hui,he told me this following maybe is true,Now How prove it?

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

Answer

Notice

$$\cosh(2nx) = T_{2n}(\cosh x)\quad\text{ and }\quad \cos(2nx) = T_{2n}(\cos x)$$

where $T_{2n}(z)$ is a
Chebyshev polynomial of the first kind. Using the $2^{nd}$ relation above, it is clear the roots of $T_{2n}(x)$ has the
form:
$$\pm\cos(\frac{2k+1}{4n}\pi),\quad\text{ for } k = 0,\ldots, n-1$$

From this, we arrive following expansion of $\cosh(2n x)$:

$$\cosh(2n x ) = A \prod_{k=0}^{n-1}\left(\cosh^2 x - \cos^2(\frac{2k+1}{4n}\pi)\right)$$

for some constant $A$ we don't care.
Taking logarithm, differentiate w.r.t $x$ and divide by $2n$ for both sides, we find:
$$\begin{align}
\tanh(2n x)
= & \frac{1}{2n} \sum_{k=0}^{n-1}\frac{ 2\sinh x\cosh x}{\cosh^2 x - \cos^2(\frac{2k+1}{4n}\pi)}\\
= & \frac{1}{n}\sum_{k=0}^{n-1}\frac{ \sinh x\cosh x}{\sinh^2 x + \sin^2(\frac{2k+1}{4n}\pi)}\\
= & \frac{1}{n}\sum_{k=0}^{n-1}\frac{ \tanh x}{\tanh^2 x + \sin^2(\frac{2k+1}{4n}\pi)(1 - \tanh^2 x)}\\
\\
= & {\Large \sum_{k=0}^{n-1}\frac{\frac{\tanh x}{n \sin^2(\frac{2k+1}{4n}\pi)}
}{1 + (\frac{1}{\sin^2(\frac{2k+1}{4n}\pi)} - 1 ) \tanh^2 x} }\\

\\
= & {\Large \sum_{k=0}^{n-1}\frac{\frac{\tanh x}{n \sin^2(\frac{2k+1}{4n}\pi)}
}{1 + \frac{\tanh^2 x}{\tan^2(\frac{2k+1}{4n}\pi)}} }\\
\end{align}$$