If α,β are acute angles and cos2α=(3cos2β-1)/(3-cos2β) than prove that tanα=√2tanβ

Steps to prove tanα=√2tanβ

We know that 

⇒cos2α=(3cos2β-1)/(3-cos2β)

⇒(1-tan^2α)/(1+tan^2α)={3-(1-tan^2β)/(1+tan^2β)-1}/{3-(1-tan^2β)/(1+tan^2β)}

⇒(1-tan^2α)/(1+tan^2α)=(2-4tan^2β)/(2+4tan^2β)

⇒(1-tan^2α)/(1+tan^2α)=(1-2tan^2β)/(1+2tan^2β)

⇒{(1-tan^2α)+(1+tan^2α)}/{(1-tan^2α)-(1+tan^2α)}={(1-2tan^2β)+(1+2tan^2β)}/{(1-2tan^2β)+(1+2tan^2β)}

⇒2/(-2tan^2α)=2/(-4tan^2β)

⇒1/(tan^2α)=1/(2tan^2β)

⇒tan^2α=rt2tan^2β

⇒L.H.S=R.H.S

Hence tanα=√2tanβ is proved.

Detail Information:-

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