For 0<ϕ<π/2, if x=∑n=0∞ cos2nϕ,y=∑n=0∞ sin2nϕ ,z=∑n=0∞ cos2nϕsin2nϕ, then
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a
xyz=xz+y
b
xyz=xy+z
c
xyz=x+y+z
d
yz=yz+x
answer is B.
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Detailed Solution
All are infinite geometric progression with common ratio < Ix=11−cos2ϕ=1sin2ϕ,y=11−sin2ϕ=1cos2ϕ,z=11−cos2ϕsin2ϕnow, xy+z=1sin2ϕcos2ϕ+11−sin2ϕcos2ϕ =1sin2ϕcos2ϕ1−sin2ϕcos2ϕxy+z=xyz-----------(i)clearly, x+y=sin2ϕ+cos2ϕsin2ϕcosϕ=xyx+y+z=xyz [using Eq. (i)]