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Question

If 0<θ,ϕ<π2, x=n=0cos2nθ, y=n=0sin2nϕ and z=n=0cos2nθsin2nϕ then :

A
xyz=4
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B
xyz=(x+y)z
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C
xy+xy+zx=z
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D
xy+z=(x+y)z
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Solution

The correct option is D xy+z=(x+y)z
x=1+cos2θ+cos4θ+
x=11cos2θ (For GP: S=a1r)x=1sin2θ (1)

y=1+sin2ϕ+sin4ϕ+
y=11sin2ϕy=1cos2ϕ (2)

z=11cos2θsin2ϕ=11(11x)(11y) [From (1) and (2)]z=xyxy(x1)(y1)

xz+yzz=xy
xy+z=(x+y)z

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