Question

HEY GUYS

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solve the above problem​

Answer 1

the given equation is: cos(θ+2α)=m.cosθ ....(1)

⇒1m=cosθcos(θ+2α)applying componendo and dividendo:1+m1−m=cosθ+cos(θ+2α)cosθ−cos(θ+2α)1+m1−m=2cosθ+(θ+2α)2.cosθ−(θ+2α)22sinθ+(θ+2α)2.sin(θ+2α)−θ21+m1−m=cos(θ+α).cosαsin(θ+α).sinα1+m1−m*sin(θ+α)cos(θ+α)=cosαsinα

⇒(1+m)(1−m)*tan(θ+α)=cotα

hope this helps you. ...pls mark it as brainliest

Answer 2

HELLO .....

the given equation is: cos(θ+2α)=m.cosθ ....(1)

⇒1m=cosθcos(θ+2α)applying componendo and dividendo:1+m1−m=cosθ+cos(θ+2α)cosθ−cos(θ+2α)1+m1−m=2cosθ+(θ+2α)2.cosθ−(θ+2α)22sinθ+(θ+2α)2.sin(θ+2α)−θ21+m1−m=cos(θ+α).cosαsin(θ+α).sinα1+m1−m*sin(θ+α)cos(θ+α)=cosαsinα

⇒(1+m)(1−m)*tan(θ+α)=cotα

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HOPE THIS HELPS YOU