We have $\log⁡\tan ⁡x=y$

(taking antilog on both side)

$∴\tan x=e^y$

L.H.S $=\sin h(n+1)y+\sin ⁡h(n-1)y$

By Forctorization formula.

$=2 \sin⁡ h[\dfrac {(n+1)y+(n-1)y}2] \cos ⁡h[\dfrac {(n+1)y-(n-1)y}2] \\ =2 \sin⁡ h[\dfrac {y(n+1+n-1)}2] \cos ⁡h[\dfrac {y(n+1-n+1)}2] \\ =2 \sin h \dfrac {2ny}2 \cos⁡ h \dfrac {2y}2 \\ =2 \sin h\space ny\space \cos⁡ h \space y \\ =2 \sin⁡ h\space n\space y [\dfrac {e^y+e^{-y}}2] \\ ∵\cos⁡ h y=\dfrac {e^y+e^{-y}}2$

L.H.S $=\sin h \space n\space y [e^y+\dfrac 1{e^y} ]\\ =\sin h\space n\space y [\tan⁡ x+\dfrac 1{\tan ⁡x} ]\\ =\sin h\space n\space y [\dfrac {\tan^2 x+1}{\tan ⁡x} ] \\ =\sin h\space n\space y [\dfrac {\dfrac 1{\cos^2⁡x}}{\sin\dfrac {x}{\cos ⁡x}} ] \\ =\sin⁡ h\space n\space y [\dfrac 1{\sin x \cos ⁡x}]×\dfrac 22 \\ =\sin h \space ny \space \dfrac 2{\sin 2x} \\ =2 \sin h\space n\space y \space cosec 2x$

=R.H.S