$cosec \bigg( \dfrac{\pi}{4} + ix \bigg) \;=\; u + iv \\ \; \\ \; \\ \therefore \dfrac{1}{sin \bigg( \dfrac{\pi}{4} + ix \bigg)} \;=\; u+iv \; \; \; \; \therefore \dfrac{1}{u+iv} \;=\; sin \bigg( \dfrac{\pi}{4} + ix \bigg) \\ \; \\ \; \\ \therefore sin\dfrac{\pi}{4} cos(ix) + cos\dfrac{\pi}{4} sin(ix) \;=\; \dfrac{u-iv}{(u+iv)(u-iv)} \; \; \; \{ \ldots Rationalising\} \\ \; \\ \; \\ \therefore \dfrac{1}{\sqrt{2}} cosh x + i \dfrac{1}{\sqrt{2}} sinh x \;=\; \dfrac{u-iv}{u^2-i^2v^2} \; \; \; \{ \because cos(ix)=coshx \; \; \& \; sin(i)=isinx \} \\ \; \\ \; \\ \therefore \dfrac{1}{\sqrt{2}} cosh x + i \dfrac{1}{\sqrt{2}} sinh x \;=\; \dfrac{u}{u^2+v^2} \; - \; i\dfrac{v}{u^2+v^2} \\ \; \\ \; \\ $

Comparing real and imaginary parts,

$ \therefore \dfrac{1}{\sqrt{2}} cosh x \;=\; \dfrac{u}{u^2+v^2} \; \; or \; \; cosh x \;=\; \dfrac{ \sqrt{2} u}{u^2+v^2} \\ \; \\ \; \\ $

On squaring, $ cosh^2 x \;=\; \dfrac{ 2 u^2}{(u^2+v^2)^2} \; \; \ldots (i) \\ \; \\ \; \\ Similarly, \; \dfrac{1}{\sqrt{2}} sinh x \;=\; \dfrac{-v}{u^2+v^2} \; \; or \; \; sinh x \;=\; \dfrac{ -\sqrt{2} u}{u^2+v^2} \\ \; \\ \; \\ $

On squaring, $ sinh^2 x \;=\; \dfrac{ 2 v^2}{(u^2+v^2)^2} \; \; \ldots (ii) \\ \; \\ \; \\ $

Equation (i) – Equation(ii)

$ \therefore cosh^2x - sinh^2x \;= \; \dfrac{2(u^2-v^2)}{(u^2+v^2)^2} \\ \; \\ \; \\ 1 \;= \; \dfrac{2(u^2-v^2)}{(u^2+v^2)^2} \; \; \; \{ \because cosh^2x - sinh^2x \;= \; 1 \} \\ \; \\ \; \\ \; \\ 2(u^2-v^2) \;= \; (u^2+v^2)^2 \; \; \; Hence \; Proved. $