For  $0<\varphi <\pi /2,$ if  $x=\sum _{n=0}^{\infty }{\cos }^{2n}\varphi ,y=\sum _{n=0}^{\infty }{\sin }^{2n}\varphi ,z=\sum _{n=0}^{\infty }{\cos }^{2n}\varphi {\sin }^{2n}\varphi$  then

For $0<\varphi <\frac{\pi }{2} x=\sum _{n=0}^{\infty }{\cos }^{2n}\varphi =1+{\cos }^2\varphi +{\cos }^4\varphi +\mathrm{....}\infty$    
 $\Rightarrow x=\frac{1}{1-\cos \varphi }=\frac{1}{{\sin }^2\varphi }\mathrm{..............}\left(i\right)$
 [using sum of infinite G.P., ${\cos }^2\alpha$  being <1]
 $y=\sum _{n=0}^{\infty }{\sin }^{2n}\varphi =1+{\sin }^2\varphi +{\sin }^4\varphi +\mathrm{........}\infty$
$\Rightarrow y=\frac{1}{1-{\sin }^2\varphi }=\frac{1}{{\cos }^2\varphi }$  ………….. (ii)
And  $z=\sum _{n=0}^{\infty }{\cos }^{2n}\varphi {\sin }^{2n}\varphi$
 $\Rightarrow z=1+{\cos }^2\varphi {\sin }^2\varphi +{\cos }^4\varphi {\sin }^4\varphi +\mathrm{......}\infty$
$\Rightarrow z=\frac{1}{1-{\cos }^2\varphi {\sin }^2\varphi }$                   …………. (iii)
On substituting the values of  ${\cos }^2\varphi$  and  ${\sin }^2\varphi$  in (iii), from 
(i) and (ii), we get 
 .$z=\frac{1}{1-\frac{1}{x}.\frac{1}{y}}\Rightarrow z=\frac{xy}{xy-1}\Rightarrow xyz-z=xy\Rightarrow xyz=xy+z$
Now,  $x+y+z=\frac{1}{{\cos }^2\varphi }+\frac{1}{{\sin }^2\varphi }+\frac{1}{1-{\cos }^2\varphi {\sin }^2\varphi }$

$$\begin{gathered} =\frac{[{\sin }^2\varphi (1-{\cos }^2\varphi {\sin }^2\varphi )+{\cos }^2\varphi (1-{\cos }^2\varphi {\sin }^2\varphi )+\left({\cos }^2\varphi {\sin }^2\varphi \right)]}{{\cos }^2\varphi {\sin }^2\varphi (1-{\cos }^2\varphi {\sin }^2\varphi )} \\ =\frac{({\sin }^2\varphi +{\cos }^2\varphi )(1-{\cos }^2\varphi {\sin }^2\varphi )+{\cos }^2\varphi {\sin }^2\varphi }{{\cos }^2\varphi {\sin }^2\varphi (1-{\cos }^2\varphi {\sin }^2\varphi )} \\ =\frac{1}{{\cos }^2\varphi {\sin }^2\varphi (1-{\cos }^2\varphi \sin 2)\varphi }=xyz \end{gathered}$$