c$ os(x+iy)cos(u+iv)=1 \\ \; \\ \; \\ \therefore cos(x+iy)=\dfrac{1}{cos(u+iv)} \;=\; sec(u+iv)\; \; \; \ldots (i) \\ \; \\ \; \\ Now, sin(x+iy) \;=\; \sqrt{1-cos^2(x+iy)} \; \; \{ \because sin^2\theta + cos^2\ theta =1 \} \\ \; \\ \; \\ =\sqrt{1-sec^2(u+iv)} \; \; \; \ldots from \; (ii) \\ \; \\ \; \\ = \sqrt{-tan^2(u+iv)} \; \; \{ \because 1 + tan^2\theta = sec^2 \theta \} \\ \; \\ \; \\ = i tan(u+iv) \\ \; \\ \; \\ sin(x+iy)= i tan(u+iv) \; \; \ldots (ii) \\ \; \\ \; \\ \therefore tan(x+iy) \;=\; \dfrac{sin(x+iy)}{cos(x+iy)} \;=\; \dfrac{itan(u+iv)}{sec(u+iv)} \; \; \; from \; (i) \; and \; (ii) \\ \; \\ \; \\ = \dfrac{isin(u+iv)}{cos(u+iv)} \times cos(u+iv) \\ \; \\ \; \\ \therefore tan(x+iy)\;=\; isin(u+iv) \; \; \; \ldots (iii) \\ \; \\ \; \\ Similarly, \\ \; \\ tan(x-iy)\;=\; isin(u-iv) \; \; \; \ldots (iv) \\ \; \\ \; \\ then \; tan2iy \;=\; tan[(x+iy)-(x-iy)] \;=\; \dfrac{tan(x+iy)-tan(x-iy)}{1+tan(x+iy)tan(x-iy)} \\ \; \\ \; \\ \therefore itanh(2y) \;=\; \dfrac{i[sin(u+iv)+sin(u-iv)]}{1+sin(u+iv)sin(u-iv)} \\ \; \\ \; \\ \therefore itanh(2y) \;=\; \dfrac{i[2sinu \; cos(iv)]}{1+\dfrac{1}{2}[cos(u+iv-u+iv)-cos(u+iv+u-iv)]} \\ \; \\ \; \\ \therefore tanh(2y) \;=\; \dfrac{2sinu \; cosh v}{1+\dfrac{1}{2}[2cosh^2v-1-1+2sin^2u]} \;=\; \dfrac{2sinu \; cosh v}{1+cosh^v-1+sin^2u} \\ \; \\ \; \\ \therefore tanh \; 2y \;=\; \dfrac{2sinu \; cosh v}{cosh^v+2sin^2u} \\ \; \\ \; \\ \therefore \dfrac{2tanh\;y}{1+tanh^2y} \;=\; \dfrac{2 sinu/ coshv}{1 + \dfrac{sin^2u}{cosh^2 v}} \; \; \ldots \{ Dividing \; num.\; and \; deno. \; By \; cosh^2 v \} \\ \; \\ \; \\ $

Comparing both the sides, we get

$ tanh^2y \;=\; \dfrac{sin^2u}{cosh^2v} \; \; or \; tan^2y cosh^2v \;=\; sin^2 u \;\;\;\;\;\;\;\; hence \; proved \\ \; \\ \; \\ \; \\ \; \\ Proving \; tan(x-iy)\;=\;-isin(u-iv) \\ \; \\ \; \\ tan(x+iy)\;=\;isin(u+iv) \\ \; \\ \; \\ \therefore \dfrac{tanx+itanhy}{1-tanxitanhy} \;=\; i[sinu.cos(iv)+cosu.sin(iv)] \\ \; \\ \; \\ \therefore \dfrac{tanx+itanhy}{1-tanx.itanhy} \;=\; i[sinu .coshv +i cosu \;sinhv] \;=\; isinu .coshv - cosu \;sinhv \\ \; \\ \; \\ \therefore tanx+itanhy \;=\; (1-tanxitanhy) ( isinu .coshv - cosu \;sinhv ) \\ \; \\ \; \\ = -cosu. sinh v \;+ \; isinu .cosh v \;+\; itanx.tany \; cosu.sinhv \;+\; (tanx .tanhy . sinu . coshv) \\ \; \\ \; \\ \therefore tanx+itanhy \;=\; [-cosh . sinhv + tanx . tany . sinu . coshv\ ] + i[tanx .tanh y. cosu. sinhv + sinu . cosh v] \\ \; \\ \; \\ $

Comparing real and imaginary parts, we get

$ tanx \;=\; -cosh \; sinhv + tanx \; tany \; sinu \; coshv \; \; and \\ \; \\ tanhy \;=\; tanx \; tanhy \; cosu \; sinhv + sinu \; cosh v \\ \; \\ \; \\ \therefore tanx - itanhy \;=\; [ -cosh \; sinhv + tanx \; tany \; sinu \; coshv ]- [ tanx \; tanhy \; cosu \; sinhv + sinu \; cosh v ] \\ \; \\ \; \\ \therefore tanx - itanhy \;=\; -cosu . sinh v(1+itanxtanhy) + sinu . coshv (tanx . tanhy - i) \\ \; \\ \; \\ = -cosu . sinhv (1+itanx\; tany) + \dfrac{sinu \; coshv}{i} \bigg( itanx \; tany - i^2 \bigg) \\ \; \\ \; \\ = -cosu . sinhv (1+itanx\; tany) + \dfrac{sinu \; coshv}{i} \bigg( 1 + itanx \; tany \bigg) \\ \; \\ \; \\ = -cosu . sinhv (1+itanx\; tany) -i{sinu \; coshv} \bigg( 1 + itanx \; tany \bigg) \\ \; \\ \; \\ \dfrac{tanx+itanhy}{(1-tanx . itanhy)} \;=\; ( -cosu . sinhv -i{sinu \; coshv} ) \;=\; -i( -icosu . sinhv +{sinu \; coshv} ) \\ \; \\ \; \\ = -i( {sinu \; coshv} -icosu . sinhv ) \\ \; \\ \; \\ = -i( {sinu \; cos(iv)} -icosu . sin(iv) ) \\ \; \\ \; \\ = -isin(u-iv) \; \; \; \; \; Hence \; Proved. $