$Q. if \ tan (x+iy)=a+ib\\ P.T\ \tanh 2y=\dfrac{2b}{1+a^2+b^2}$

$Here\ we \ can\ use\ this \ formuala\\ tan(A-B)=\dfrac{ tanA-tanB}{1+tanA.tanB}$

$\tan 2iy =tan[(x+iy)-(x-iy)]\\ =\dfrac{\tan(x+iy)-\tan(a-ib)}{1+tan(x+iy)(tan(x-iy)}\\ tan(x+iy)=a+ib, tan(x-iy)=a+ib \ and \tan i2y=i\tanh2y \\ substituting\ the\ value\ in \ the\ above\ equation\\ $

$i\tanh2y =\dfrac{(a+ib)-(a-ib)}{1+(a+ib)(a-ib)}\\ i\tanh2y= \dfrac{i2b}{1+a^2+b^2}\\ \tanh2y= \dfrac{2b}{1+a^2+b^2} (Proved)$