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Q.

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

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a

xy+z

b

xz+y

c

x+y+z

d

yz+x

answer is A, C.

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Detailed Solution

x=1+cos2ϕ+cos4ϕ+........=11cos2ϕ=1sin2ϕsin2ϕ=1x

y=1+sin2ϕ+sin4ϕ......=11sin2ϕ=1cos2ϕcos2ϕ=1y

Similarly    z=11cos2ϕsin2ϕ=111x1y=xyxy1

xyzz=xy

xyz=xy+z

xy=1sin2ϕ.1cos2ϕ=sin2ϕ+cos2ϕsin2ϕcos2ϕ=x+y

xyz=x+y+z

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