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Question Number 94882 by i jagooll last updated on 21/May/20

solve the equation tan θ+tan 2θ+tan 3θ = tan θ.tan 2θ.tan 3θ

$$\mathrm{solve}\:\mathrm{the}\:\mathrm{equation}\: \\ $$$$\mathrm{tan}\:\theta+\mathrm{tan}\:\mathrm{2}\theta+\mathrm{tan}\:\mathrm{3}\theta\:=\: \\ $$$$\mathrm{tan}\:\theta.\mathrm{tan}\:\mathrm{2}\theta.\mathrm{tan}\:\mathrm{3}\theta\: \\ $$

Answered by john santu last updated on 21/May/20

we have identity tan 3θ.tan 2θ.tan θ= tan 3θ−tan 2θ−tan θ then we get tan 3θ−tan 2θ−tan θ= tan 3θ+tan 2θ+tan θ ⇒2tan 2θ+2tan θ=0 ((2tan θ)/(1−tan^2 θ)) + tan θ = 0 tan θ (3−tan^2 θ)=0 { ((tan θ=0)),((tan θ=(√3))),((tan θ=−(√3))) :} now it easy to solve

$$\mathrm{we}\:\mathrm{have}\:\mathrm{identity}\: \\ $$$$\mathrm{tan}\:\mathrm{3}\theta.\mathrm{tan}\:\mathrm{2}\theta.\mathrm{tan}\:\theta= \\ $$$$\mathrm{tan}\:\mathrm{3}\theta−\mathrm{tan}\:\mathrm{2}\theta−\mathrm{tan}\:\theta \\ $$$$\mathrm{then}\:\mathrm{we}\:\mathrm{get}\:\mathrm{tan}\:\mathrm{3}\theta−\mathrm{tan}\:\mathrm{2}\theta−\mathrm{tan}\:\theta= \\ $$$$\mathrm{tan}\:\mathrm{3}\theta+\mathrm{tan}\:\mathrm{2}\theta+\mathrm{tan}\:\theta \\ $$$$\Rightarrow\mathrm{2tan}\:\mathrm{2}\theta+\mathrm{2tan}\:\theta=\mathrm{0} \\ $$$$\frac{\mathrm{2tan}\:\theta}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \theta}\:+\:\mathrm{tan}\:\theta\:=\:\mathrm{0} \\ $$$$\mathrm{tan}\:\theta\:\left(\mathrm{3}−\mathrm{tan}\:^{\mathrm{2}} \theta\right)=\mathrm{0} \\ $$$$\begin{cases}{\mathrm{tan}\:\theta=\mathrm{0}}\\{\mathrm{tan}\:\theta=\sqrt{\mathrm{3}}}\\{\mathrm{tan}\:\theta=−\sqrt{\mathrm{3}}}\end{cases} \\ $$$$\mathrm{now}\:\mathrm{it}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve} \\ $$

Commented by i jagooll last updated on 21/May/20

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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