If $0<\varphi <\frac{\pi }{2},$ and $x=\sum _{n=0}^{\mathrm{\infty }} {\cos }^{2n} \varphi ,$

$y=\sum _{n=0}^{\mathrm{\infty }} {\sin }^{2n} \varphi$ $z=\sum _{n=0}^{\mathrm{\infty }} {\cos }^{2n} \varphi {\sin }^{2n} \varphi ,$ 

We have $x=\frac{1}{1-{\cos }^2 \varphi }=\frac{1}{{\sin }^2 \varphi }$ and $y=\frac{1}{{\cos }^2 \varphi }$

$\begin{array}{l}\text{ Also, }z=\frac{1}{1-{\cos }^2\varphi {\sin }^2\varphi }=\frac{1}{1-\frac{1}{xy}}=\frac{xy}{xy-1}\\ \Rightarrow xyz-z=xy\text{ or }xyz=xy+z\\ \text{ But }\frac{1}{x}+\frac{1}{y}=1\Rightarrow x+y=xy\\ \therefore \end{array}$