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Question

Suppose θ and (ϕ0) are such that sec(θ+ϕ), secθ and sec(θϕ) are in A.P. If cosθ=kcos(ϕ2) for some k, then k is equal to?

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Solution

sec(θ+ϕ)+sec(θϕ)=2secθ
cos(θ+ϕ)+cos(θϕ)cos(θ+ϕ)cos(θϕ)=2cosθ
2cosθcosϕcos2θcos2ϕ=1cosθ
cos2θcosϕ=cos2θ+cos2ϕ1
cos2θ(cosϕ1)=cos2ϕ1
cos2θ=cosϕ+1
cos2θ=2cos2ϕ2
cosθ=2cosϕ2
k=2

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