t$an \bigg( \dfrac{\pi}{4} + iv \bigg) \;=\; \dfrac{ tan\dfrac{\pi}{4} + tan(iv) }{ 1-tan\dfrac{\pi}{4}tan(iv) } \\ \; \\ \; \\ = \dfrac{1+itanv}{1-itanv} \\ \{ \because tan(A+B)= \dfrac{ tanA + tanB }{ 1-tanA tanB } \\ and \; tan(ix)=itanx \} \\ \; \\ \; \\ \dfrac{1+i \dfrac{sinhv}{coshv}}{1-i\dfrac{sinhv}{coshv}} \;=\; = \dfrac{coshv+isinhv}{coshv-isinhv} \\ \; \\ \; \\ \; \\ tan \bigg( \dfrac{\pi}{4} + iv \bigg) \;=\; \dfrac{(coshv+isinhv)(coshv+isinhv)}{(coshv-isinhv) (coshv+isinhv) } \\ \; \\ \; \\ \; \\ = \dfrac{ cosh^2v + i^2sinh^2v+2isinhvcoshv }{ cosh^v - i^2sinh^2v } \\ \; \\ \; \\ = \dfrac{ cosh^2v - sinh^2v+2isinhvcoshv }{ cosh^v + sinh^2v } \; \; \; \ldots \{ \because i^2=-1 \} \\ \; \\ \; \\ = \dfrac{1+i(sinh2v)}{cosh2v} \; \; \{ cosh^2-sinh^2=1 \; \& \; cosh^2+sinh^2=cosh2v \} \\ \; \\ \; \\ = sech2v+i(tanh2v) \\ \; \\ \; \\ \; \\ But, \; tan \bigg( \dfrac{\pi}{4} + iv \bigg) \;=\; re^{i\theta} \;=\; r(cos\theta+isin\theta) \\ \; \\ \; \\ = rcos\theta + i(rsin\theta) \\ \; \\ \; \\ \therefore rcos\theta + i(rsin\theta) \; =\; sech2v+i(tanh2v) \\ \; \\ $

Comparing real and imaginary parts,

$ \\ \; \\ \; \\ \; \\ (i) \\ \; \\ \therefore rcos\theta = sech2v \; \; \therefore r^2cos^2\theta \;= \; sech^22v \; \; \ldots(i) \\ \; \\ \therefore rsin\theta = tanh2v \; \; \therefore r^2sin^2\theta \;= \; tanh^22v \; \; \ldots(ii) \\ \; \\ $

Adding (i) and (ii),

$ r^2(sin^2\theta + cos^2\theta) \;=\; sech^2 2v + tanh^2 2v \\ \; \\ \; \\ \; \\ \therefore r^2(1) = 1 \; \; \; \{ sin^2\theta + cos^2\theta \; \; sech^2 2v = 1 -tanh^2x \} \\ \; \\ \; \\ \therefore r^2=1 \\ \; \\ \; \\ \therefore r=1 \; \; \; \; \; \; Hence \; Proved. \\ \; \\ \; \\ \; \\ (ii) \\ \; \\ rsin\theta= tanh2v \; and \; rcos\theta = sech2v \\ \; \\ \therefore \dfrac{rsin\theta}{r cos\theta} \;=\; \dfrac{sinh(2v)/cosh(2v)}{1/cosh(2v)} \\ \; \\ \therefore tan\theta \;=\; sinh2v \; \; \; Hence Proved. \\ \; \\ \; \\ \; \\ (iii) \\ \; \\ tanh2v=sin\theta \; \; \{ \because r=1\} \\ \; \\ \therefore \dfrac{2tanhv}{1+tanh^2v} \;=\; sin\theta \;=\; 2sin\dfrac{\theta}{2}cos\dfrac{\theta}{2} \;=\; \\ \; \\ \therefore \dfrac{2tanhv}{1+tanh^2v} \;=\; \dfrac{2 \dfrac{sin \theta/2}{cos \theta/2}}{ sech^2 \theta/2} \\ \; \\ = \dfrac{2tan \theta/2}{1 + tanh^2 \theta/2} \\ \; \\ \; \\ On \; comparing \; both \; sides, \; we \; get \; tanhv \;=\; tan \dfrac{\theta}{2} \; \; \; \; \ldots Hence \; Proved. $