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Q.

If 0<θ,ϕ<π2,x=n=0cos2nθ,y=n=0sin2nϕandz=n=0cos2nθsin2nϕthen

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a

xyz =4

b

xy – z =(x + y)z

c

xy + yz + zx =z

d

xy + z = (x + y)z

answer is D.

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Detailed Solution

x=1+cos2θ+cos4θ+.....=11cos2θ=1sin2θ

y=1+sin2ϕ+sin4ϕ+......=11sin2ϕ=1cos2ϕ

Z=1+cos2θsin2ϕ+(cos2θsin2ϕ)2+.....=11cos2θsin2ϕ

Z=11(11/x)(11/y)=xyx+y1

z(x+y)z=xyz(x+y)=xy+z

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