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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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