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Question Number 20647 by oyshi last updated on 30/Aug/17

prove it, tan (α+(π/3))+tan (α−(π/3))=((4sin 2α)/(1−4sin^2 α))

$${prove}\:{it}, \\ $$$$\mathrm{tan}\:\left(\alpha+\frac{\pi}{\mathrm{3}}\right)+\mathrm{tan}\:\left(\alpha−\frac{\pi}{\mathrm{3}}\right)=\frac{\mathrm{4sin}\:\mathrm{2}\alpha}{\mathrm{1}−\mathrm{4sin}\:^{\mathrm{2}} \alpha} \\ $$

Answered by $@ty@m last updated on 30/Aug/17

LHS=((tanα+tan(π/3))/(1−tanα.tan(π/3)))+((tanα−tan(π/3))/(1+tanα.tan(π/3))) =((tanα+(√3))/(1−(√3).tanα))+((tanα−(√3))/(1+(√3).tanα)) =(((1+(√3).tanα)(tanα+(√3))+(1−(√3).tanα)(tanα−(√3)))/((1−(√3).tanα)(1+(√3).tanα))) =((tanα+(√3)tan^2 α+3tanα+(√3)+tanα−(√3)tan^2 α−(√3)+3tanα)/(1−3tan^2 α)) =((8tanα)/(1−3tan^2 α))=((4sin2α)/(cos^2 α−3sin^2 α)) (multiplying N^r & D^r by cos^2 α) =((4sin2α)/(1−sin^2 α−3sin^2 α)) =((4sin 2α)/(1−4sin^2 α)) =RHS

$${LHS}=\frac{{tan}\alpha+{tan}\frac{\pi}{\mathrm{3}}}{\mathrm{1}−{tan}\alpha.{tan}\frac{\pi}{\mathrm{3}}}+\frac{{tan}\alpha−{tan}\frac{\pi}{\mathrm{3}}}{\mathrm{1}+{tan}\alpha.{tan}\frac{\pi}{\mathrm{3}}} \\ $$$$=\frac{{tan}\alpha+\sqrt{\mathrm{3}}}{\mathrm{1}−\sqrt{\mathrm{3}}.{tan}\alpha}+\frac{{tan}\alpha−\sqrt{\mathrm{3}}}{\mathrm{1}+\sqrt{\mathrm{3}}.{tan}\alpha} \\ $$$$=\frac{\left(\mathrm{1}+\sqrt{\mathrm{3}}.{tan}\alpha\right)\left({tan}\alpha+\sqrt{\mathrm{3}}\right)+\left(\mathrm{1}−\sqrt{\mathrm{3}}.{tan}\alpha\right)\left({tan}\alpha−\sqrt{\mathrm{3}}\right)}{\left(\mathrm{1}−\sqrt{\mathrm{3}}.{tan}\alpha\right)\left(\mathrm{1}+\sqrt{\mathrm{3}}.{tan}\alpha\right)} \\ $$$$=\frac{{tan}\alpha+\sqrt{\mathrm{3}}{tan}^{\mathrm{2}} \alpha+\mathrm{3}{tan}\alpha+\sqrt{\mathrm{3}}+{tan}\alpha−\sqrt{\mathrm{3}}{tan}^{\mathrm{2}} \alpha−\sqrt{\mathrm{3}}+\mathrm{3}{tan}\alpha}{\mathrm{1}−\mathrm{3}{tan}^{\mathrm{2}} \alpha} \\ $$$$=\frac{\mathrm{8}{tan}\alpha}{\mathrm{1}−\mathrm{3}{tan}^{\mathrm{2}} \alpha}=\frac{\mathrm{4}{sin}\mathrm{2}\alpha}{{cos}^{\mathrm{2}} \alpha−\mathrm{3}{sin}^{\mathrm{2}} \alpha} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({multiplying}\:{N}^{{r}} \:\&\:{D}^{{r}} \:{by}\:{cos}^{\mathrm{2}} \alpha\right) \\ $$$$=\frac{\mathrm{4}{sin}\mathrm{2}\alpha}{\mathrm{1}−{sin}^{\mathrm{2}} \alpha−\mathrm{3}{sin}^{\mathrm{2}} \alpha} \\ $$$$=\frac{\mathrm{4sin}\:\mathrm{2}\alpha}{\mathrm{1}−\mathrm{4sin}\:^{\mathrm{2}} \alpha} \\ $$$$={RHS} \\ $$

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