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Q.

For0<θ<π2 , If x=n=0cos2nϕy=n=0sin2nϕ and z=n=0cos2nϕsin2nϕ then xyz=

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a

xy+z

b

yz+x

c

x+y+z

d

xz+y

answer is A, C.

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Detailed Solution

x=n=0cos2nϕ=1sin2ϕsin2ϕ=1x

y=n=0sin2nϕ=1cos2ϕcos2ϕ=1y

z=n=0cos2nϕsin2nϕ=11cos2ϕsin2ϕ=111x(1y)

z=xyxy1xyzz=xy

xyz=xy+z

sin2ϕ+cos2ϕ=1x+1y1=y+xxy

xy=x+y

xyz=xy+z=x+y+z

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