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Question

If tan(3π11)+4sin(2π11)=λ then the value of λ29 is equal to ___

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Solution

λ=1cos(3π14){sin(3π11)+4sin(2π11)cos(3π11)}λcos(3π11)=sin(3π11)+4sin(2π11)cos(3π11)λ2cos2(3π11)=sin2(3π11)+16sin2(2π11)cos2(3π11)+8sin(2π11)sin(3π11)cos(3π11)λ2(2cos2(3π11))=2sin2(3π11)+32sin2(2π11)cos2(3π11)+8sin(2π11)sin(6π11)

λ2(2cos2(3π11))=(1cos(6π11))+8(1cos4π11)(1+cos6π11)+4(cos4π11cos8π11)

λ2(2cos2(3π11))=9+7cos(6π11)4cos(4π11)8cos(4π11)cos(6π11)4cos(8π11)

λ2(2cos2(3π11))=9+7cos(6π11)4cos(4π11)4(cos(10π11)+cos(2π11))4cos(8π11)

λ2(2cos2(3π11))=9+11cos(6π11)4{cos(2π11)+cos(4π11)+cos(6π11)+cos(8π11)+cos(10π11)}

λ2(2cos2(3π11))=9+11cos(6π11)4(sinπsinπ112sinπ11)

λ2(2cos2(3π11))=9+11cos(6π11)+2

λ2(2cos2(3π11))=11(1+cos(6π11))
λ2(2cos2(3π11))=11(2cos2(311))

λ2=11
λ29=119=2


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