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Question Number 73560 by mhmd last updated on 13/Nov/19

$$provethat(1-itan\theta )/(i+cot\theta )=itan\theta$$$$pleassirhelpme?$$

Commented by MJS last updated on 13/Nov/19

$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{generally}\mathrm{true}$$$$\frac{1-\mathrm{i}\tan \theta }{\mathrm{i}+\cot \theta }=2\sin \theta \cos \theta -\tan \theta -2\mathrm{i}{\sin }^2\theta$$$${\mathrm{we}}^{\prime }\mathrm{d}\mathrm{have}\mathrm{to}\mathrm{have}$$$$2\sin \theta \cos \theta -\tan \theta =0\wedge -{2\sin }^2\theta =\tan \theta$$$$\mathrm{which}\mathrm{is}\mathrm{only}\mathrm{true}\mathrm{for}$$$$\theta =2n\pi \vee \theta =\frac{8n-1}{4}\pi \vee \theta =\frac{8n+3}{4}\pi$$

Answered by ajfour last updated on 13/Nov/19

$$\frac{1-i\tan \theta }{i+\cot \theta }=\frac{(1-i\tan \theta )\left(\tan \theta \right)}{1+i\tan \theta }$$$$={(1-i\tan \theta )}^2\sin \theta \cos \theta$$$$=e^{-2i\theta }\tan \theta$$$$..$$

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