Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com