$\displaystyle Let\ f(z)=\ u+iv\ be\ analytic\ \\[2ex] \displaystyle By\ Cauchy\ rehmans\ equations \\[2ex] Ux=Vy\ and\ Uy=-Vx\ \grave{a}(1)Given\ a=\frac{\sin{2x}}{\cosh{2y}-\cos{2x}} \displaystyle Let\ D=\ \cosh{2y}-\cos{2x} \\[2ex] \displaystyle Differentiating\ wrt\ x, \\[2ex] $

$\displaystyle \therefore{}Ux=(\cosh{2y}-\cos{2x})\bullet{}\cos{2x}-\sin{2x}\bullet{}(0+\sin{2x}\bullet{}2)/\ D^2 \\[2ex] \displaystyle =2(\cosh{2y}\cos{2x}\ {\cos}^2{2y}-\ {\sin}^2{2x}/D^2\ \\[2ex] $

$ =-\ 2\sin{2x}.\sinh{2y}/\ D^2...(3)\\[2ex]\therefore{}f’(z)=Ux-iUy %eq15 \\[2ex] \displaystyle =2(\cosh{2y}\cos{2x}-1)/{\ (\cosh{2y}-\cos{2x})}^2+i\bullet{}2\sin{2x}\bullet{}\sinh{2y}/\ {\ (\cosh{2y}-\cos{2x})}^2 \\[2ex] $

$\displaystyle By\ Milne\ Thompsons\ Method,\ \\[2ex] \displaystyle Put\ x=z,\ and\ y=0, \\[2ex] \displaystyle f'(z)=\frac{2(1\bullet{}\cos{2z}-1)}{{(1-\cos{2z)}}^2}+0 \\[2ex] \displaystyle =\frac{-2(1-\cos{2z}) }{{(1-\cos{2z)}}^2} \\[2ex] \displaystyle =\frac{-2}{(1-\cos{2z})} \\[2ex] \displaystyle =-\ {csc}^2z \\[2ex] $

$\displaystyle \therefore{} f(z)=\int f ^{'}\left(z\right)dz \\[2ex] \displaystyle =-\int \ {csc}^2zdz \\[2ex] \displaystyle =\ \cot z+k \\[2ex] $