If x=sum_{n=0}^{infty} cos ^{2 n} theta,y=sum | vedclass

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(A) Given $x = \sum_{n=0}^{\infty} \cos^{2n} 	heta = \frac{1}{1 - \cos^2 	heta} = \frac{1}{\sin^2 	heta} \implies \sin^2 	heta = \frac{1}{x}$.Give

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$x z+y z=x y+z$

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(A) Given $x = \sum_{n=0}^{\infty} \cos^{2n} 	heta = \frac{1}{1 - \cos^2 	heta} = \frac{1}{\sin^2 	heta} \implies \sin^2 	heta = \frac{1}{x}$.Given $y = \sum_{n=0}^{\infty} \sin^{2n} 	heta = \frac{1}{1 - \sin^2 	heta} = \frac{1}{\cos^2 	heta} \implies \cos^2 	heta = \frac{1}{y}$.Since $\sin^2 	heta + \cos^2 	heta = 1$,we have $\frac{1}{x} + \frac{1}{y} = 1$,which implies $x + y = xy$.Given $z = \sum_{n=0}^{\infty} (\cos^2 	heta \sin^2 	heta)^n = \frac{1}{1 - \cos^2 	heta \sin^2 	heta} = \frac{1}{1 - \frac{1}{x} \cdot \frac{1}{y}} = \frac{xy}{xy - 1}$.From $xy - 1 = \frac{xy}{z}$,we have $xy - 1 = \frac{x+y}{z}$ (since $xy = x+y$).Thus,$z(xy - 1) = x + y \implies xyz - z = x + y$.Since $x+y = xy$,we can write $xyz - z = xy$,or $xyz = xy + z$.Alternatively,using $xy = x+y$,we have $z(x+y) = xy + z$,which simplifies to $xz + yz = xy + z$.

If $x=\sum_{n=0}^{\infty} \cos ^{2 n} \theta$,$y=\sum_{n=0}^{\infty} \sin ^{2 n} \theta$,$z=\sum_{n=0}^{\infty} \cos ^{2 n} \theta \sin ^{2 n} \theta$ and $0 < \theta < \frac{\pi}{2}$,then

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