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Question Number 63763 by aliesam last updated on 08/Jul/19

Commented by Prithwish sen last updated on 08/Jul/19

(2/(1+cos(2x) + i sin(2x))) = ((2{1+cos(2x) − isin(2x)})/(1+2cos(2x)+cos^2 (2x) + sin^2 (2x))) = ((1+cos(2x) − isin(2x))/(1+ cos(2x))) = 1 − i((2sinx.cosx)/(2cos^2 x)) = 1 −itanx Hence proved.

$$\frac{\mathrm{2}}{\mathrm{1}+\mathrm{cos}\left(\mathrm{2x}\right)\:+\:\mathrm{i}\:\mathrm{sin}\left(\mathrm{2x}\right)}\:=\:\frac{\mathrm{2}\left\{\mathrm{1}+\mathrm{cos}\left(\mathrm{2x}\right)\:−\:\mathrm{isin}\left(\mathrm{2x}\right)\right\}}{\mathrm{1}+\mathrm{2cos}\left(\mathrm{2x}\right)+\mathrm{cos}^{\mathrm{2}} \left(\mathrm{2x}\right)\:+\:\mathrm{sin}^{\mathrm{2}} \left(\mathrm{2x}\right)} \\ $$$$=\:\frac{\mathrm{1}+\mathrm{cos}\left(\mathrm{2x}\right)\:−\:\mathrm{isin}\left(\mathrm{2x}\right)}{\mathrm{1}+\:\mathrm{cos}\left(\mathrm{2x}\right)}\:=\:\mathrm{1}\:−\:\mathrm{i}\frac{\mathrm{2sinx}.\mathrm{cosx}}{\mathrm{2cos}^{\mathrm{2}} \mathrm{x}} \\ $$$$=\:\mathrm{1}\:−\mathrm{itanx}\:\mathrm{Hence}\:\mathrm{proved}. \\ $$

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