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Question

Prove that tanα+tanβ+tanγtanαtanβtanγ=sin(α+β+γ)cosαcosβcosγ

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Solution

sin(α+β+γ)
=sin(α+β)cosγ+cos(α+β)sinγ
=sinαcosβcosγ+cosαsinβcosγ+cosαcosβsinγsinαsinβsinγ
Now divide each term by cosαcosβcosγ
Then Lhs=tanα+tanβ+tanγtanαtanβtanγ
LHS=RHS
Hence proved.

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