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x=sum(n=0)^(oo)cos^(2n)theta, y=sum(n=0)...

`x=sum_(n=0)^(oo)cos^(2n)theta, y=sum_(n=0)^(oo)sin^(2n)theta, z=sum_(n=0)^(oo)cos^(2n)sin^(2n)theta, |cos theta| lt 1, |sin theta| lt 1` then `x+y+z` is equal to

A

xy

B

yz

C

zx

D

xyz

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The correct Answer is:
d
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