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Question

If sin(θ+ϕ)=nsin(θϕ),n1, then the value of tanθtanϕ is equal to

A
nn1
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B
n+1n1
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C
n1n
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D
n1n+1
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E
1+n1n
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Solution

The correct option is B n+1n1
sin(θ+ϕ)=nsin(θϕ)
sin(θ+ϕ)sin(θϕ)=n
Applying componendo and dividendo, we get
sin(θ+ϕ)+sin(θϕ)sin(θ+ϕ)sin(θϕ)=n+1n1
[2sin(θ+ϕ+θϕ2)cos(θ+ϕθ+ϕ2)][2cos(θ+ϕ+θϕ2)sin(θ+ϕθ+ϕ2)]=n+1n1
sinθcosϕcosθsinϕ=n+1n1
tanθtanϕ=n+1n1

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