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Question Number 12078 by agni5 last updated on 12/Apr/17
If tan^2 α = 1+2tan^2 β then prove that cos 2β =1+2cos 2α .
$$\mathrm{If}\:\mathrm{tan}\:^{\mathrm{2}} \alpha\:=\:\mathrm{1}+\mathrm{2tan}\:^{\mathrm{2}} \beta\:\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\:\mathrm{cos}\:\mathrm{2}\beta\:=\mathrm{1}+\mathrm{2cos}\:\mathrm{2}\alpha\:. \\ $$
Answered by ajfour last updated on 12/Apr/17
1+2cos 2α =1+((2(1−tan^2 α))/(1+tan^2 α)) =((1+tan^2 α+2−2tan^2 α)/(1+tan^2 α)) =((3−tan^2 α)/(1+tan^2 α)) =((3−(1+2tan^2 β))/(1+(1+2tan^2 β))) =((2(1−tan^2 β))/(2(1+tan^2 β))) = ((1−tan^2 β)/(1+tan^2 β)) = cos 2β .
$$\mathrm{1}+\mathrm{2cos}\:\mathrm{2}\alpha\:=\mathrm{1}+\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \alpha\right)}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \alpha} \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \alpha+\mathrm{2}−\mathrm{2tan}\:^{\mathrm{2}} \alpha}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \alpha} \\ $$$$\:\:\:\:=\frac{\mathrm{3}−\mathrm{tan}\:^{\mathrm{2}} \alpha}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \alpha}\:=\frac{\mathrm{3}−\left(\mathrm{1}+\mathrm{2tan}\:^{\mathrm{2}} \beta\right)}{\mathrm{1}+\left(\mathrm{1}+\mathrm{2tan}\:^{\mathrm{2}} \beta\right)} \\ $$$$\:\:\:\:=\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \beta\right)}{\mathrm{2}\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \beta\right)}\:=\:\frac{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \beta}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \beta} \\ $$$$\:\:\:=\:\mathrm{cos}\:\mathrm{2}\beta\:. \\ $$

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