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Question

Prove that:

tanθ tan(60θ) tan(60+θ)


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    Solution

    We have,
    LHS=tanθ tan(60θ) tan(60+θ)=sinθ(60θ)sin(60+θ)cosθ cos(60θ)cos(60+θ)=2sinθ sin(60θ)sin(60θ)cosθ cos(60θ)cos(60+θ)=sinθ[2sin(60θ)sin(60+θ)]cosθ[2cos(60θ)cos(60+θ)]=sinθ[cos{(60θ)(60+θ)}cos{(60θ)+(60+θ)}]cosθ[cos{(60θ)+(60+θ)}]+[cos{(60θ)(60+θ)}]=sinθ[cosθ(2θ)cos120]cosθ[cos120+cos(2θ)]=sinθ[cos2θcos120]cosθ[cos120+cos2θ] [cos(θ)=cosθ]=sinθ[cos2θ+sin30]cos[cos(90+30)+cos2θ]=sinθ[cos2θ+sin30]cosθ[sin30+cos2θ] [ cos is negative in IInd quadrant]
    =sinθ[cos2θ+12]cos[12+cos2θ]=sinθ cos2θ+12sinθ12cosθ+cosθ cos2θ=sinθ cos2θ+12sinθcosθ cos2θ12cosθ=12[2sinθ cos2θ]+12sinθ12[2cosθ cos2θ]12cosθ=12[2sinθ cos2θ+sinθ]12[2cosθ cos2θcosθ]=[sin3θ+sin(θ)+sinθ][cos3θ+cos)θ]cosθ=sin3θsinθ+sinθcos3θ+cosθcosθ[ sin(θ)=sinθ and cos(θ)=cosθ]=sin3θcos3θ=tan3θ=RHStanθ tan(60θ)tan(60+θ)=tan3θ.


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