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Question

# If sin θ+sin ϕ=a and cos θ+cos ϕ=b, (a≠b, a≠0, b≠0) then -

A
tan θ+tan ϕ=8ab(a2+b2)2+4b2
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B
cos θ.cos ϕ=(a2+b2)24a24(a2+b2)
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C
cos (θ+ϕ)=b2a2b2+a2
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D
sin (θ+ϕ)=4ab(a2+b2)+2b2
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Solution

## The correct option is C cos (θ+ϕ)=b2−a2b2+a2Squaring and adding 1+1+2 cos(θ−ϕ)=a2+b2⇒cos(θ−ϕ)=a2+b2−22 2 sin(θ+ϕ2) cos(θ−ϕ2)=a2 cos(θ+ϕ2) cos(θ−ϕ2)=b–––––––––––––––––––––––––––––tan(θ+ϕ2)=ab cos(θ+ϕ)=1−tan2(θ+ϕ2)1+tan2(θ+ϕ2)=1−a2b21+a2b2=(b2−a2b2+a2) sin(θ+ϕ)=2 tan(θ+ϕ2)1+tan2(θ+ϕ2)=2×ab1+a2b2=(2aba2+b2) cos θ cos ϕ=12(cos(θ+ϕ)+cos(θ−ϕ))=12(b2−a2b2+a2+a2+b2−22) =(a2+b2)2−4a24(a2+b2) tanθ+tanϕ=sinθcosθ+sinϕcosϕ=sin(θ+ϕ)cosθ.cosϕ=8ab(a2+b2)2−4a2

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