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Question

If tanx+tan(x+π3)+tan(x+2π3)=3, prove that 3tan xtan3x13tan2 x=1.

Or

If sinθ=nsin(θ+2α), prove that tan(θ+α)=1+n1ntanα.

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Solution

we have ,tanx+tan(x+π3)+tan(x+2π3)=3

tanx+tan(x+π3)+tan(x+ππ3)=3

tanx+tan(x+π3)+tan{π+(xπ3)}=3

tanx+tan(x+π3)+(xπ3)=3

[tan(π+θ)=tanθ] tanx+tanx+tanπ31tanx.tanπ3+tan xtanπ31+tan x.π3

[tan(A+B)=tanA+taB1tanA tanBand tan(AB)=tanAtanB1+tanA tanB]

[tanx3tan3x+tan x+3tan2x+3+3tan x+tan x3tan2x3+3tan x]13tan2x=3

9tan xtan3 x13tan2 x=33tan xtan3x13tan2x=1

Hence proved

Or

We have, sinθ=nsin(θ+2α)

sin(θ+2α)sinθ=1n

sin(θ+2α)+sinθsinθ=1+n1n

[using componendo and dividendo rule]

2sin(θ+2α+θ)2cos(θ+2αθ)22cos(θ+2α+θ)2sin(θ+2αθ)2=1+n1n

[sin C+sinD=2sin(C+D2)cos(CD2)and sin Csin D=2cos(C+D2)sin(CD2)]

2sin(θ+α)cosα2cos(θ+α)sinα=1+n1n

tan(θ+α)cotα=1+n1n

tan(θ+α)=1+n1ntanα Hence proved.


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