If $$0 < \theta, \phi < \frac{\pi}{2}$$, $$x = \sum_{n=0}^{\infty} \cos^{2n}\theta$$, $$y = \sum_{n=0}^{\infty} \sin^{2n}\phi$$ and $$z = \sum_{n=0}^{\infty} \cos^{2n}\theta \cdot \sin^{2n}\phi$$ then:

Since $$0 < \theta, \phi < \frac{\pi}{2}$$, we have $$0 < \cos^2\theta < 1$$ and $$0 < \sin^2\phi < 1$$. Each given series is a geometric series.

For $$x$$: $$x = \sum_{n=0}^{\infty} \cos^{2n}\theta = \frac{1}{1 - \cos^2\theta} = \frac{1}{\sin^2\theta}$$, so $$\cos^2\theta = 1 - \frac{1}{x}$$.

For $$y$$: $$y = \sum_{n=0}^{\infty} \sin^{2n}\phi = \frac{1}{1 - \sin^2\phi} = \frac{1}{\cos^2\phi}$$, so $$\sin^2\phi = 1 - \frac{1}{y}$$.

For $$z$$: $$z = \sum_{n=0}^{\infty} (\cos^2\theta \cdot \sin^2\phi)^n = \frac{1}{1 - \cos^2\theta \cdot \sin^2\phi}$$.

Now we compute $$\cos^2\theta \cdot \sin^2\phi = \left(1 - \frac{1}{x}\right)\left(1 - \frac{1}{y}\right) = 1 - \frac{1}{x} - \frac{1}{y} + \frac{1}{xy}$$.

Therefore $$z = \frac{1}{\frac{1}{x} + \frac{1}{y} - \frac{1}{xy}} = \frac{xy}{x + y - 1}$$.

Cross-multiplying: $$z(x + y - 1) = xy$$, which gives $$zx + zy - z = xy$$, and rearranging yields $$xy + z = z(x + y)$$.

Therefore, the correct relation is $$xy + z = (x + y)z$$.