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If $\cos α + \cos β = a, \sin α + \sin β = b$ and $α - β = 2 θ,$ then $\frac{\cos 3 θ}{\cos θ} =$

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a2+b2−2

a2+b2−3

3−a2−b2

a2+b2/4

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Correct option is B

From the given relations we havea2+b2=cos2⁡α+cos2⁡β+2cos⁡αcos⁡β+sin2⁡α+sin2⁡β+2sin⁡αsin⁡β=2+2cos⁡(α−β)=2+2cos⁡2θ=4cos2⁡θNow cos⁡3θcos⁡θ=4cos3⁡θ−3cos⁡θcos⁡θ=4cos2⁡θ−3=a2+b2−3.

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If cos⁡α+cos⁡β=a,sin⁡α+sin⁡β=b and α−β=2θ, then cos⁡3θcos⁡θ=

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If $\cos α + \cos β = a, \sin α + \sin β = b$ and $α - β = 2 θ,$ then $\frac{\cos 3 θ}{\cos θ} =$