If 0<θ<π2,x=∑n=0∞cos2nθ,y=∑n=0∞sin2nθ, and z=∑n=0∞cos2nθsin2nθ, then xyz
xyz=xz+y
xyz=xy+z
xyz=yz+x
none of these
Explanation for correct option:
Step-1: Simplify x=∑n=0∞cos2nθ.
Given, x=∑n=0∞cos2nθ
⇒ x=1+cos2θ+cos4θ+cos6θ+.......(are in G.P)
⇒ x=11-cos2θ∵S∞=a1-r
⇒ x=1sin2θ.........i
Step-2: Simplify y=∑n=0∞sin2nθ.
Given, y=∑n=0∞sin2nθ
⇒ y=1+sin2θ+sin4θ+sin6θ+.......(are in G.P)
⇒ y=11-sin2θ∵S∞=a1-r
⇒ y=1cos2θ.........ii
Step-3: Simplify z=∑n=0∞cos2nθsin2nθ
⇒ z=∑n=0∞cos2nθsin2nθ
⇒ z=1+cosθsinθ+cos2θsin2θ+cos3θsin3θ+.........(are in G.P)
⇒ z=11-sin2θcos2θ..........iii
Step-4: Solving equation i,ii&iii.
⇒ z=11-1xy
⇒ z=xyxy-1
⇒ xy=xyz-z
⇒ xyz=xy+z
Hence correct answer is option B.