If x=sum_{n=0}^{infty}(-1)^{n} tan ^{2 n} the | vedclass

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(C) The given series for $x$ is a geometric series with first term $a=1$ and common ratio $r=-	an^2 	heta$.
Since $0 < 	heta < \frac{\pi}{4}

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$y(1-x)=1$

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(C) The given series for $x$ is a geometric series with first term $a=1$ and common ratio $r=-	an^2 	heta$.
Since $0 < 	heta < \frac{\pi}{4}$,we have $0 < 	an^2 	heta < 1$,so the series converges to $x = \frac{1}{1 - (-	an^2 	heta)} = \frac{1}{1 + 	an^2 	heta} = \frac{1}{\sec^2 	heta} = \cos^2 	heta$.
The given series for $y$ is a geometric series with first term $a=1$ and common ratio $r=\cos^2 	heta$.
Since $0 < 	heta < \frac{\pi}{4}$,we have $\frac{1}{2} < \cos^2 	heta < 1$,so the series converges to $y = \frac{1}{1 - \cos^2 	heta} = \frac{1}{\sin^2 	heta}$.
From the expression for $x$,we have $\cos^2 	heta = x$,which implies $\sin^2 	heta = 1 - x$.
Substituting this into the expression for $y$,we get $y = \frac{1}{1 - x}$.
Rearranging this gives $y(1 - x) = 1$.

If $x=\sum_{n=0}^{\infty}(-1)^{n} \tan ^{2 n} \theta$ and $y=\sum_{n=0}^{\infty} \cos ^{2 n} \theta,$ for $0 < \theta < \frac{\pi}{4},$ then

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