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

$$\begin{gathered} \begin{array}{l}x=1-(\tan \theta {)}^2+(\tan \theta {)}^4\dots \dots \infty \\ x=\frac{1}{1+{\tan }^2\theta }\Rightarrow x=\frac{1}{{\sec }^2\theta }={\cos }^2\theta \\ y=(\cos \theta {)}^0+(\cos \theta {)}^2+(\cos \theta {)}^4\dots \dots \infty \\ y=\frac{1}{1-{\cos }^2\theta }=\frac{1}{{\sin }^2\theta }\\ \frac{1}{y}={\sin }^2\theta \\ {\cos }^2\theta +{\sin }^2\theta =1\\ \Rightarrow x+\frac{1}{y}=1\Rightarrow xy+1=y \\ \Rightarrow xy-y=-1\Rightarrow y(1-x)=1\end{array} \end{gathered}$$