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Q.

x=∑n=0∞ cos2n⁡θ,y=∑n=0∞ sin2n⁡θ,z=∑n=0∞ cos2n⁡θsin2n⁡θ|cos⁡θ|<1, |sin⁡θ|<1 then x+y+z is equal to

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a

xy

b

yz

c

zx

d

xyz

answer is D.

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Detailed Solution

x=11−cos2⁡x=1sin2⁡x,y=1cos2⁡x and z=11−sin2⁡xcos2⁡x=11−1x⋅1y=xyxy−1⇒z(xy−1)=xy=x+y⇒x+y+z=xyz.
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x=∑n=0∞ cos2n⁡θ,y=∑n=0∞ sin2n⁡θ,z=∑n=0∞ cos2n⁡θsin2n⁡θ|cos⁡θ|<1, |sin⁡θ|<1 then x+y+z is equal to