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Question

# If tan(3π11)+4sin(2π11)=λ then the value of λ2−9 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))=(1−cos(6π11))+8(1−cos4π11)(1+cos6π11)+4(cos4π11−cos8π11) 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(sinπ−sinπ112sinπ11) λ2(2cos2(3π11))=9+11cos(6π11)+2 λ2(2cos2(3π11))=11(1+cos(6π11)) λ2(2cos2(3π11))=11(2cos2(311)) λ2=11 ∴λ2−9=11−9=2

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