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Q.

For 0<θ<π2, if x=n-0Cos2nθ,y=n-0sin2nθ and  z=n-0Cos2nθsin2n θ then

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a

xyz = yz + x

b

xyz = xy - z

c

xyz = x + y + z

d

xyz = xz + y

answer is C.

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Detailed Solution

x=n=0 cos2 nθ =1+cos2θ+cos4θ+..... =11-cos2θ=1sin2θsin2θ=1xy=n=0 sin2nθ =1+sin2θ+sin4θ+.....

 =11-sin2θ=1cos2θ  cos2θ=1y, 1x+1y=1    xy=1z=r=0 sin2rθ cos2rθ =1+sin2θ cos2θ+sin4θ cos4θ+..... =11-sin2θ cos2θr=11-1x·1yz=xyxy-1  xyz-z=xyxyz=xy+zxyz=x+y+z

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