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Question: For 0 \<θ\<π/2, if x = \(\sum_{n = 0}^{\infty}{\cos^{2n}\theta,}y = \sum_{n = 0}^{\infty}{\sin^{2n}...

For 0 <θ<π/2, if x =

n=0cos2nθ,y=n=0sin2nθ,z=n=0cos2nθsin2nθ,\sum_{n = 0}^{\infty}{\cos^{2n}\theta,}y = \sum_{n = 0}^{\infty}{\sin^{2n}\theta,}z = \sum_{n = 0}^{\infty}{\cos^{2n}\theta\sin^{2n}\theta,} then

A

xyz = xz + y

B

xyz = xy +z

C

xyz = yz +x

D

None of these

Answer

xyz = xy +z

Explanation

Solution

Here x = 11cos2θ\frac{1}{1 - \cos^{2}\theta} = cosec2θ , y = sec2 θ ,

z = 11sin2θcos2θ\frac{1}{1 - \sin^{2}\theta\cos^{2}\theta}

z =111x.1y=xyxy1\frac{1}{1 - \frac{1}{x}.\frac{1}{y}} = \frac{xy}{xy - 1} ⇒ xyz = xy + z