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Question Number 54543 by Tawa1 last updated on 05/Feb/19

$$\mathrm{Prove}\mathrm{that}:\frac{{\mathrm{z}}^2-1}{{\mathrm{z}}^2+1}=\mathrm{i}\tan \left(\theta \right)$$$$\mathrm{where}\mathrm{z}=\cos \left(\theta \right)+\mathrm{i}\sin \left(\theta \right)$$

Commented by maxmathsup by imad last updated on 06/Feb/19

$$wehavez=e^{i\theta }\Rightarrow \frac{z^2-1}{z^2+1}=\frac{e^{2i\theta }-1}{e^{2i\theta }+1}=\frac{e^{i\theta }-e^{-i\theta }}{e^{i\theta }+e^{-i\theta }}=i\frac{\frac{e^{i\theta }-e^{-i\theta }}{2i}}{\frac{e^{i\theta }+e^{-i\theta }}{2}}=i\frac{sin\theta }{cos\theta }=itan\theta$$

Answered by tanmay.chaudhury50@gmail.com last updated on 05/Feb/19

$$\frac{z^2-1}{z^2+1}$$$$=\frac{z-\frac{1}{z}}{z+\frac{1}{z}}$$$$z=cos\theta +isin\theta =e^{i\theta }$$$$\frac{1}{z}=\frac{1}{e^{i\theta }}=e^{-i\theta }=cos(-\theta )+isin(-\theta )=cos\theta -isin\theta$$$$so\frac{z-\frac{1}{z}}{z+\frac{1}{z}}=\frac{2isin\theta }{2cos\theta }=itan\theta$$

Commented by Tawa1 last updated on 05/Feb/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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