If x=∑n=0∞(-1)n tan2nθ and y=∑n=0∞ cos2nθ for θ<θ<π4, then:

x=∑n=0∞ (-1)n tan2nθ =1-tan2θ+tan4θ =11+tan2θ =1sec2θy=∑n=0∞ cos2nθ =1+cos2θ+cos4θ+......y=11-cos2θy=11-xy(1-x)=1

If x=∑n=0∞(-1)n tan2nθ and y=∑n=0∞ cos2nθ for θ<θ<π4, then:

x=∑n=0∞ (-1)n tan2nθ =1-tan2θ+tan4θ =11+tan2θ =1sec2θy=∑n=0∞ cos2nθ =1+cos2θ+cos4θ+......y=11-cos2θy=11-xy(1-x)=1

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

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$x (1 + y) = 1$

$x (1 - y) = 1$

$y (1 - y) = 1$

$y (1 + x) = 1$

answer is D.

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Detailed Solution

$\begin{array}{rcl} x & = & \sum_{n = 0}^{\infty} {(- 1)}^{n} \tan^{2 n} θ \\ = & 1 - \tan^{2} θ + \tan^{4} θ \\ = & \frac{1}{1 + \tan^{2} θ} \\ = & \frac{1}{\sec^{2} θ} \\ y & = & \sum_{n = 0}^{\infty} \cos^{2 n} θ \\ = & 1 + \cos^{2} θ + \cos^{4} θ + . . . . . . \\ y & = & \frac{1}{1 - \cos^{2} θ} \\ y & = & \frac{1}{1 - x} \\ y (1 - x) & = & 1 \end{array}$