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Q.

If 0<θ<π2 and x=n=0cos2nθ,

y=n=0sin2nθ,

z=n=0cos2nθsin2nθ,

then

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a

xyz=xz+y

b

xyz=xy+z

c

xyz=yz+x

d

xy+yz+zx=1

answer is B.

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Detailed Solution

x=1+cos2θ+cos4θ+....,11cos2θ=1sin2θ similarity y=1cos2θandz=11cos2θsin2θ,1sin2θcos2θ+11sin2θcos2θ xyz=1sin2θcos2θ(1cos2θsin2θ) 1sin2θcos2θ+11sin2θcos2θ=xy+z

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