tan theta + tan 2 theta +√3 tan theta tan2theta=√3​

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Here is the solution of your asked query:
tanθ+tan2θ+3tanθ tan2θ=3tanθ+tan2θ=3-3tanθ tan2θtanθ+tan2θ=31-tanθ tan2θtanθ+tan2θ1-tanθ tan2θ=3tanθ+2θ = tanπ3                      {since tanA+tanB1-tanA tanB=tanA+B}θ+2θ=π3  3θ=π3θ=π9
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