$tan(\theta+i\phi)=tan\alpha+isec\alpha\\ tan(\theta-i\phi)=tan\alpha-isec\alpha\\ tan 2\theta=tan[(\theta+i\phi)+(\theta-i\phi)]\\ =\dfrac{tan(\theta+i\phi)+tan(\theta-i\phi)}{1-tan(\theta+i\phi)tan(\theta-i\phi) }\\ =\dfrac{tan\alpha+isec\alpha+tan\alpha-isec\alpha}{1-(tan\alpha+isec\alpha)(tan\alpha-isec\alpha) }\\ =\dfrac{2tan\alpha}{1-(tan^2\alpha+sec^2\alpha)} =-\dfrac{tan\alpha}{tan^2\alpha}=-cot\alpha= tan(\pi/2+\alpha)\\ \therefore 2\theta=n\pi+\dfrac\pi 2+\alpha$

$tan(\theta+i\phi)=tan\alpha+isec\alpha\\ tan(\theta-i\phi)=tan\alpha-isec\alpha\\ tan i2\phi=tan[(\theta+i\phi)-(\theta-i\phi)] =\dfrac{tan(\theta+i\phi)-tan(\theta-i\phi)}{1+tan(\theta+i\phi)tan(\theta-i\phi) }\\ =\dfrac{tan\alpha+isec\alpha-tan\alpha+isec\alpha}{1-(tan\alpha+isec\alpha)(tan\alpha-isec\alpha) }\\ =\dfrac{2isec\alpha}{1-(tan^2\alpha+sec^2\alpha)} =i\dfrac{sec\alpha}{sec^2\alpha}=icos\alpha\\ tan i2\phi=itanh2\phi=icos\alpha\\ 2\phi=tanh^{-1}cos\alpha\bigg(tanh^{-1}z=\dfrac 12log\bigg[\dfrac{1+z}{1-z}\bigg]\bigg)$

$\therefore2\phi=\dfrac 12log\bigg[\dfrac{1+cos\alpha}{1-cos\alpha}\bigg]\\ =\dfrac 12log\bigg[\dfrac{cos^2\alpha/2}{sin^2\alpha/2}\bigg]=log cot(\ alpha2)\\ \therefore e^{2\phi}= cot\bigg(\dfrac{\alpha}2\bigg)$