If α and β are acute angles and cos2α=3cosβ−13−cos2β, then prove that tanα=√2tanβ.
I tried this question by taking the formula of cos2α in terms of tan (which is of degree two) but I couldn't prove it. Please suggest some hints.
Answer
Using Weierstrass substitution in either sides.
1−tan2α1+tan2α=3⋅1−tan2β1+tan2β−13−1−tan2β1+tan2β
⟹1−tan2α1+tan2α=2−4tan2β2+4tan2β
Using Componendo and dividendo, tan2α1=4tan2β2
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