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Question-75209

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Question Number 75209 by vishalbhardwaj last updated on 08/Dec/19
Answered by Kunal12588 last updated on 08/Dec/19
x+y+z=xyz let x=tan α,y=tan β, z=tan γ tan α + tan β + tan γ = tan α tan β tan γ ⇒tan γ (1−tan α tan β)=−(tan α+tan β) ⇒tan γ=−((tan α+tanβ)/(1−tan α tan β))=−tan (α+β) ⇒tan (α+β)=−tan γ = tan (π−γ) ⇒α+β=π−γ ⇒α+β+γ=π ⇒2α+2β=2π−2γ ⇒tan(2α+2β)=tan(2π−2γ) ⇒((tan 2α + tan 2β)/(1−tan 2α tan 2β))=−tan 2γ ⇒tan 2α+tan 2β =−tan 2γ+tan 2α tan 2β tan 2γ ⇒tan 2α+tan 2β+tan2γ=tan 2α tan 2β tan 2γ ⇒((2tan α)/(1−tan^2 α))+((2tan β)/(1−tan^2 β))+((2tan γ)/(1−tan^2 γ))=((2tan α)/(1−tan^2 α))×((2tan β)/(1−tan^2 β))×((2tan γ)/(1−tan^2 γ)) ⇒((2x)/(1−x^2 ))+((2y)/(1−y^2 ))+((2z)/(1−z^2 ))=(((2x)/(1−x^2 )))(((2y)/(1−y^2 )))(((2z)/(1−z^2 )))
$${x}+{y}+{z}={xyz} \\ $$$${let}\:{x}={tan}\:\alpha,{y}={tan}\:\beta,\:{z}={tan}\:\gamma \\ $$$${tan}\:\alpha\:+\:{tan}\:\beta\:+\:{tan}\:\gamma\:=\:{tan}\:\alpha\:{tan}\:\beta\:{tan}\:\gamma \\ $$$$\Rightarrow{tan}\:\gamma\:\left(\mathrm{1}−{tan}\:\alpha\:{tan}\:\beta\right)=−\left({tan}\:\alpha+{tan}\:\beta\right) \\ $$$$\Rightarrow{tan}\:\gamma=−\frac{{tan}\:\alpha+{tan}\beta}{\mathrm{1}−{tan}\:\alpha\:{tan}\:\beta}=−{tan}\:\left(\alpha+\beta\right) \\ $$$$\Rightarrow{tan}\:\left(\alpha+\beta\right)=−{tan}\:\gamma\:=\:{tan}\:\left(\pi−\gamma\right) \\ $$$$\Rightarrow\alpha+\beta=\pi−\gamma \\ $$$$\Rightarrow\alpha+\beta+\gamma=\pi \\ $$$$\Rightarrow\mathrm{2}\alpha+\mathrm{2}\beta=\mathrm{2}\pi−\mathrm{2}\gamma \\ $$$$\Rightarrow{tan}\left(\mathrm{2}\alpha+\mathrm{2}\beta\right)={tan}\left(\mathrm{2}\pi−\mathrm{2}\gamma\right) \\ $$$$\Rightarrow\frac{{tan}\:\mathrm{2}\alpha\:+\:{tan}\:\mathrm{2}\beta}{\mathrm{1}−{tan}\:\mathrm{2}\alpha\:{tan}\:\mathrm{2}\beta}=−{tan}\:\mathrm{2}\gamma \\ $$$$\Rightarrow{tan}\:\mathrm{2}\alpha+{tan}\:\mathrm{2}\beta\:=−{tan}\:\mathrm{2}\gamma+{tan}\:\mathrm{2}\alpha\:{tan}\:\mathrm{2}\beta\:{tan}\:\mathrm{2}\gamma \\ $$$$\Rightarrow{tan}\:\mathrm{2}\alpha+{tan}\:\mathrm{2}\beta+{tan}\mathrm{2}\gamma={tan}\:\mathrm{2}\alpha\:{tan}\:\mathrm{2}\beta\:{tan}\:\mathrm{2}\gamma \\ $$$$\Rightarrow\frac{\mathrm{2}{tan}\:\alpha}{\mathrm{1}−{tan}^{\mathrm{2}} \alpha}+\frac{\mathrm{2}{tan}\:\beta}{\mathrm{1}−{tan}^{\mathrm{2}} \beta}+\frac{\mathrm{2}{tan}\:\gamma}{\mathrm{1}−{tan}^{\mathrm{2}} \gamma}=\frac{\mathrm{2}{tan}\:\alpha}{\mathrm{1}−{tan}^{\mathrm{2}} \alpha}×\frac{\mathrm{2}{tan}\:\beta}{\mathrm{1}−{tan}^{\mathrm{2}} \beta}×\frac{\mathrm{2}{tan}\:\gamma}{\mathrm{1}−{tan}^{\mathrm{2}} \gamma} \\ $$$$\Rightarrow\frac{\mathrm{2}{x}}{\mathrm{1}−{x}^{\mathrm{2}} }+\frac{\mathrm{2}{y}}{\mathrm{1}−{y}^{\mathrm{2}} }+\frac{\mathrm{2}{z}}{\mathrm{1}−{z}^{\mathrm{2}} }=\left(\frac{\mathrm{2}{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\right)\left(\frac{\mathrm{2}{y}}{\mathrm{1}−{y}^{\mathrm{2}} }\right)\left(\frac{\mathrm{2}{z}}{\mathrm{1}−{z}^{\mathrm{2}} }\right) \\ $$
Answered by Kunal12588 last updated on 08/Dec/19

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