1. Identify both infinite series as geometric series

Given $x=\sum_{n=0}^{\infty}(-1)^n\tan^{2n}\theta$ and $y=\sum_{n=0}^{\infty}\cos^{2n}\theta$ with $0<\theta<\frac{\pi}{4}$.

Since $0<\theta<\frac{\pi}{4}$, we have:

• $0<\tan\theta<1 \implies 0<\tan^2\theta<1$
• $0<\cos\theta<1 \implies 0<\cos^2\theta<1$

So both are convergent geometric series.

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2. Find $x$

$x=1-\tan^2\theta+\tan^4\theta-\tan^6\theta+\cdots$ This is a geometric series with first term $1$ and common ratio $-\tan^2\theta$.

Hence, $x=\frac{1}{1-(-\tan^2\theta)}=\frac{1}{1+\tan^2\theta}$ Using the identity $1+\tan^2\theta=\sec^2\theta,$ we get $x=\frac{1}{\sec^2\theta}=\cos^2\theta.$

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3. Find $y$

$y=1+\cos^2\theta+\cos^4\theta+\cdots$ This is a geometric series with first term $1$ and common ratio $\cos^2\theta$.

Thus, $y=\frac{1}{1-\cos^2\theta}$ Using $1-\cos^2\theta=\sin^2\theta,$ we get $y=\frac{1}{\sin^2\theta}=\csc^2\theta.$

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4. Check each option

Option A: $x(1+y)=1$

Substitute $x=\cos^2\theta$, $y=\csc^2\theta$: $x(1+y)=\cos^2\theta\left(1+\csc^2\theta\right)$ This is not identically $1$. So, A is false.

Option B: $y(1-x)=1$

Now, $1-x=1-\cos^2\theta=\sin^2\theta$ Therefore, $y(1-x)=\csc^2\theta\cdot \sin^2\theta=1$ So, B is true.

Option C: $y(1+x)=1$

$y(1+x)=\csc^2\theta(1+\cos^2\theta)$ This is not identically $1$. So, C is false.

Option D: $x(1-y)=1$

$x(1-y)=\cos^2\theta(1-\csc^2\theta)$ This is not identically $1$. So, D is false.

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5. Final answer

The correct option is: $\boxed{\text{B}}$

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6. Comparison with stored answer

Stored correct answer: B

My derived answer also is B, so they agree.