<s>[INST] Finding the relation between function x,y,z - trigo problem Problem : 
For $\displaystyle 0 < \theta < \frac{\pi}{2}$ if 
$$\begin{align}x &= \sum^{\infty}_{n =0} \cos^{2n}\theta \\ y &= \sum^{\infty}_{n =0} \sin^{2n}\theta\\ z &= \sum^{\infty}_{n =0} \cos^{2n}\theta \sin^{2n}\theta \end{align}$$
then 
options are : 
(a) $xyz = xz+y$
(b) $xyz = xy+z$
(c)  $xy^2 =y^2+x$
I have solution of this however I have one doubt in that, it is mention that : 
$\displaystyle x=\sum^{\infty}_{n =0} \cos^{2n}\theta = \frac{1}{1-\cos^2\theta}$ (How it is derived... or what about this result.). Please guide on this... thanks. 
 [/INST] $$\begin{align}x &= \sum^{\infty}_{n =0} \cos^{2n}\theta \\ y &= \sum^{\infty}_{n =0} \sin^{2n}\theta\\ z &= \sum^{\infty}_{n =0} \cos^{2n}\theta \sin^{2n}\theta \end{align}$$
$$x=1+{\cos^2\theta}+{\cos^4\theta}+\cdots\infty \;terms$$
this is infinite geometric series :
$S_{\infty}=\dfrac {a}{1-r}\;\;\,|r|<1$ ,here $\cos\theta<1, when\; 0<\theta<\dfrac \pi2 \;so\; \cos^2\theta<1$
$$x=\dfrac {1}{1-\cos^2\theta}\implies \dfrac {1}{\sin^2\theta} \implies x=\csc^2\theta$$ here $a=1$ and r=$\cos^2\theta$
similarly for y $a=1$ and $r=\sin^2\theta$
$$y=\dfrac {1}{1-\sin^2\theta}\implies \dfrac {1}{\cos^2\theta}\implies y=\sec^2\theta$$
and $$z=\dfrac {1}{1-\cos^2\theta\cdot\sin^2\theta}$$
so option (b) $$xy+z=\csc^2\theta\cdot\sec^2\theta+\dfrac {1}{1-\cos^2\theta\cdot\sin^2\theta}$$
$$xy+z=\dfrac {\csc^2\theta\cdot\sec^2\theta}{1-\cos^2\theta\cdot\sin^2\theta}$$
$$xy+z=xyz$$
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