$(i) \ cosec \left ( \dfrac {\pi}{4}+ ix \right )= u+iv $ $= \sin \left ( \dfrac {\pi}{4}+ ix \right )= \dfrac {1}{u+iv} = \dfrac {u-iv}{u^2+v^2} $ $ \therefore \sin \dfrac {\pi}{4}\cos ix+ \cos \dfrac {\pi}{4}\sin ix = \dfrac {u-iv}{u^2 +v^2} $ $\dfrac {1}{\sqrt{2}}\cosh x+ \dfrac {1}{\sqrt{2}}i \sinh x= \dfrac {u}{u^2+v^2}- \dfrac {iv}{u^2+v^2} $ Comparing real and imaginary parts, $\dfrac {1}{\sqrt{2}}\cosh x= \dfrac {u}{u^2+v^2}, \dfrac {1}{\sqrt{2}}\sinh x= - \dfrac {v}{u^2+v^2} $ $\cosh x= \dfrac {\sqrt{2}u}{u^2+v^2}, \sinh x = \dfrac {\sqrt{2}u}{u^2 + v^2}$ $\cosh^2 x- \sinh^2 x=1$ $\dfrac {2u^2}{(u^2+v^2)^2}- \dfrac {2v^2}{(u^2+v^2)^2}=1 $ $\dfrac {2 (u^2 - v^2)}{(u^2 + v^2)^2}= 1 $ $(u^2+v^2)^2=2 (u^2-v^2)$