**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} - …