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

prove that (1−itanθ)/(i+cotθ)=itanθ pleas sir help me?

$${prove}\:{that}\:\left(\mathrm{1}−{itan}\theta\right)/\left({i}+{cot}\theta\right)={itan}\theta \\ $$$${pleas}\:{sir}\:{help}\:{me}? \\ $$

Commented by MJS last updated on 13/Nov/19

it′s not generally true ((1−i tan θ)/(i+cot θ))=2sin θ cos θ −tan θ −2i sin^2 θ we′d have to have 2sin θ cos θ −tan θ =0 ∧ −2sin^2 θ =tan θ which is only true for θ=2nπ∨θ=((8n−1)/4)π∨θ=((8n+3)/4)π

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

Answered by ajfour last updated on 13/Nov/19

((1−itan θ)/(i+cot θ))=(((1−itan θ)(tan θ))/(1+itan θ)) =(1−itan θ)^2 sin θcos θ =e^(−2iθ) tan θ ..

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

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