**To prove if $\ln{\tan{x}}= y$ 

$ \sinh{\left(n+1\right)y} + \sinh{\left(n-1\right)y} = 2 \sinh{ny}{cosec{2x}}$** 

Now, as $\ln{\tan{x}}= y$ we can write,

$ e^{y}=\tan{x}$,

Also by defination of hyperbolic functions

$\sinh{x }= \dfrac{ e^{x} - e^{-x}}{2} $

Hence,

L.H.S. = $ \sinh{\left(n+1\right)y} + \sinh{\left(n-1\right)y}$

$\\ = \dfrac{ e^{\left(n+1\right)y} - e^{-\left(n+1\right)y}}{2} +\dfrac{ e^{\left(n-1\right)y} - e^{-\left(n-1\right)y}}{2} \\ = \dfrac{ e^{ny}e^{y}-e^{-ny}e^{-y} }{2} +\dfrac{ e^{ny}e^{-y}-e^{-ny}e^{y}}{2} \\ = e^{ny}\left( \dfrac{e^{y}+e^{-y}}{2}\right) - e^{-ny}\left(\dfrac{e^{y}+e^{-y}}{2}\right) \\ = \left( e^{y}+e^{-y} \right) \left( \dfrac{e^{ny}-e^{-ny}}{2} \right) \\ = \left( \tan{x} + \cot{x} \right) \sinh{ny} $

$\\ = \left( \dfrac{\sin{x}}{\cos{x}}+ \dfrac{\cos{x}}{\sin{x}} \right)\sinh{ny} \\ = \left( \dfrac{\sin^2{x}+ \cos^2{x}}{\sin{x}\cos{x}} \right)\sinh{ny} \\ = \left( \dfrac{1}{\dfrac{\sin{2x}}{2}} \right)\sinh{ny} \\ = \left( \dfrac{2}{\sin{2x}} \right)\sinh{ny} \\ = 2\sinh{ny}cosec{2x}$

$=R.H.S$

Hence Proved.