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Question

Prove that :
tan2αtan2β=sin(α+β)sin(αβ)cos2αcos2β

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Solution

Simplify the LHS of tan2αtan2β=sin(α+β)sin(αβ)cos2αcos2β

tan2αtan2β=(tanαtanβ)(tanα+tanβ)

=(sinαcosαsinβcosβ)(sinαcosα+sinβcosβ)

=(sinαcosβcosαsinβcosαcosβ)(sinαcosβ+cosαsinβcosαcosβ)

=(sin(αβ)cosαcosβ)(sin(α+β)cosαcosβ)

=sin(αβ)sin(α+β)cos2αcos2β

=RHS


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