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Question Number 143097 by ZiYangLee last updated on 10/Jun/21

$$\mathrm{If}z=\cos \theta +i\sin \theta ,$$$$\mathrm{prove}\mathrm{that}\frac{1-z^2}{1+z^2}=-i\tan \theta$$

Answered by MJS_new last updated on 10/Jun/21

$$\frac{1-{(\mathrm{c}+\mathrm{is})}^2}{1+{(\mathrm{c}+\mathrm{is})}^2}=\frac{1-{\mathrm{c}}^2+{\mathrm{s}}^2-2\mathrm{ics}}{1-{\mathrm{s}}^2+{\mathrm{c}}^2+2\mathrm{ics}}=\frac{{\mathrm{s}}^2-\mathrm{ics}}{{\mathrm{c}}^2+\mathrm{ics}}=$$$$=\frac{\mathrm{s}(\mathrm{s}-\mathrm{ic})}{\mathrm{c}(\mathrm{c}+\mathrm{is})}=\frac{-\mathrm{is}z}{\mathrm{c}z}=-\mathrm{i}\tan \theta$$

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