prove-that-tan-2tan2-4tan-4-8cot8-cot-

Question Number 43467 by peter frank last updated on 11/Sep/18

prove that tanθ+2tan2θ+4tan 4θ +8cot8θ=cotθ

$${prove}\:{th}\mathrm{at}\:\mathrm{tan}\theta+\mathrm{2}{tan}\mathrm{2}\theta+\mathrm{4}{tan}\:\:\:\mathrm{4}\theta \\ $$$$+\mathrm{8cot8}\theta={cot}\theta \\ $$

Answered by ajfour last updated on 11/Sep/18

8cot 8θ+4tan 4θ+2tan 2θ+tan θ = ((8(1−tan^2 4θ))/(2tan 4θ))+4tan 4θ+... = 4cot 4θ+2tan 2θ+tan θ = ((4(1−tan^2 2θ))/(2tan 2θ))+2tan 2θ+tan θ = 2cot 2θ+tan θ = ((2(1−tan^2 θ))/(2tan θ))+tan θ = cot θ .

$$\mathrm{8cot}\:\mathrm{8}\theta+\mathrm{4tan}\:\mathrm{4}\theta+\mathrm{2tan}\:\mathrm{2}\theta+\mathrm{tan}\:\theta \\ $$$$\:\:\:\:=\:\frac{\mathrm{8}\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \mathrm{4}\theta\right)}{\mathrm{2tan}\:\mathrm{4}\theta}+\mathrm{4tan}\:\mathrm{4}\theta+… \\ $$$$\:\:\:\:\:=\:\mathrm{4cot}\:\mathrm{4}\theta+\mathrm{2tan}\:\mathrm{2}\theta+\mathrm{tan}\:\theta \\ $$$$\:\:\:\:=\:\frac{\mathrm{4}\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \mathrm{2}\theta\right)}{\mathrm{2tan}\:\mathrm{2}\theta}+\mathrm{2tan}\:\mathrm{2}\theta+\mathrm{tan}\:\theta \\ $$$$\:\:\:=\:\mathrm{2cot}\:\mathrm{2}\theta+\mathrm{tan}\:\theta \\ $$$$\:\:\:=\:\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \theta\right)}{\mathrm{2tan}\:\theta}+\mathrm{tan}\:\theta\:\:=\:\mathrm{cot}\:\theta\:. \\ $$

Commented by peter frank last updated on 11/Sep/18

$${thanks}\: \\ $$

Commented by peter frank last updated on 11/Sep/18

sir the third line am not understood how you get 4cot 4θ

$$ \\ $$$$\mathrm{sir}\:\mathrm{the}\:\mathrm{third}\:\mathrm{line}\:\mathrm{am}\:\mathrm{not}\:\mathrm{understood} \\ $$$$\mathrm{how}\:\mathrm{you}\:\mathrm{get}\:\mathrm{4cot}\:\mathrm{4}\theta \\ $$$$ \\ $$$$ \\ $$

Commented by peter frank last updated on 11/Sep/18

Commented by ajfour last updated on 11/Sep/18

$$\frac{\mathrm{8}}{\mathrm{2tan}\:\mathrm{4}\theta}\:=\:\mathrm{4cot}\:\mathrm{4}\theta \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Sep/18

tanθ−cotθ =tanθ−(1/(tanθ)) =((tan^2 θ−1)/(tanθ)) =((−2(1−tan^2 θ))/(2tanθ)) =−2cot2θ so tanθ=cotθ−2cot2θ 2tan2θ=2(cot2θ−2cot4θ) 4tan4θ= 4(cot4θ−2cot8θ) 8cot8θ=8cot8θ now add them RHS =cotθ new tricks...no need to calculatd tan2θ, tan4θ cot8θ etc...

$${tan}\theta−{cot}\theta \\ $$$$={tan}\theta−\frac{\mathrm{1}}{{tan}\theta} \\ $$$$=\frac{{tan}^{\mathrm{2}} \theta−\mathrm{1}}{{tan}\theta} \\ $$$$=\frac{−\mathrm{2}\left(\mathrm{1}−{tan}^{\mathrm{2}} \theta\right)}{\mathrm{2}{tan}\theta} \\ $$$$=−\mathrm{2}{cot}\mathrm{2}\theta \\ $$$${so}\:\:{tan}\theta={cot}\theta−\mathrm{2}{cot}\mathrm{2}\theta \\ $$$$\:\:\:\:\:\:\mathrm{2}{tan}\mathrm{2}\theta=\mathrm{2}\left({cot}\mathrm{2}\theta−\mathrm{2}{cot}\mathrm{4}\theta\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{4}{tan}\mathrm{4}\theta=\:\:\mathrm{4}\left({cot}\mathrm{4}\theta−\mathrm{2}{cot}\mathrm{8}\theta\right) \\ $$$$\:\:\:\:\:\:\:\:\mathrm{8}{cot}\mathrm{8}\theta=\mathrm{8}{cot}\mathrm{8}\theta \\ $$$${now}\:{add}\:{them}\:\:{RHS}\:={cot}\theta\: \\ $$$${new}\:{tricks}…{no}\:{need}\:{to}\:{calculatd}\:{tan}\mathrm{2}\theta,\:\:{tan}\mathrm{4}\theta \\ $$$$\:{cot}\mathrm{8}\theta\:{etc}… \\ $$$$ \\ $$$$ \\ $$

Commented by peter frank last updated on 11/Sep/18

$${thanks} \\ $$

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