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Q.

For 0<ϕ<π2, If x=n=0cos2nϕ,y=n=0sin2nϕ and z=n=0cos2nϕsin2nϕ , then

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a

xyz = xz + y

b

xyz = xy + y

c

xyz = x + y + z

d

xyz = yz + x

answer is C.

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Detailed Solution

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we have 

x=n=0cos2nϕ=1+cos2ϕ+cos4ϕ+=11cos2ϕ=1sin2ϕsimilarly y=11sin2ϕ=1cos2ϕand z=11sin2ϕcos2xz=111x1y=xyxy1xyz=xy+z

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