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Question Number 186126 by Mingma last updated on 01/Feb/23

Commented by Frix last updated on 01/Feb/23

tan 2α =((2tan α)/(1−tan^2  α)) ⇒  tan 4α =tan (2×(2α)) =((4(1−tan^2  α)tan α)/(1−6tan^2  α +tan^4  α))

$$\mathrm{tan}\:\mathrm{2}\alpha\:=\frac{\mathrm{2tan}\:\alpha}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\alpha}\:\Rightarrow \\ $$$$\mathrm{tan}\:\mathrm{4}\alpha\:=\mathrm{tan}\:\left(\mathrm{2}×\left(\mathrm{2}\alpha\right)\right)\:=\frac{\mathrm{4}\left(\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\alpha\right)\mathrm{tan}\:\alpha}{\mathrm{1}−\mathrm{6tan}^{\mathrm{2}} \:\alpha\:+\mathrm{tan}^{\mathrm{4}} \:\alpha} \\ $$

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