• # question_answer                                             If $0<\theta ,\phi <\pi /2$ and $x=\sum\limits_{n=0}^{\infty }{{{\sin }^{2n}}\theta },y=\sum\limits_{n=0}^{\infty }{{{\cos }^{2n}}\phi }$ and $z=\sum\limits_{n=0}^{\infty }{{{\cos }^{n}}\left( \theta +\phi \right){{\cos }^{n}}\left( \theta +\phi \right)},$ then A) $xyz+1=yz-zx$ B) $xyz-1=yz+zx$ C) $xyz-xy=yz-zx$ D) $xyz+1=yz+zx$

[c]  Given, $x=\sum\limits_{n=0}^{\infty }{{{\sin }^{2n}}\theta }$ $=1+{{\sin }^{2}}\theta +{{\sin }^{4}}\theta +...$ $\Rightarrow$$x=\frac{1}{1-{{\sin }^{2}}\theta }=\frac{1}{{{\cos }^{2}}\theta }$ $\Rightarrow$$y=\sum\limits_{n=0}^{\infty }{{{\cos }^{2n}}\phi }$$=1+{{\cos }^{2}}\phi +{{\cos }^{4}}\phi +...$ $\Rightarrow y=\frac{1}{1-{{\cos }^{2}}\phi }=\frac{1}{{{\sin }^{2}}\phi }$
 $z=\sum\limits_{n=0}^{\infty }{{{\cos }^{n}}\left( \theta +\phi \right)}{{\cos }^{n}}\left( \theta +\phi \right)$ $\Rightarrow$$z=1+\cos (\theta +\phi ){{\cos }^{2}}(\alpha -\phi )$$+{{\cos }^{2}}(\theta +\phi ){{\cos }^{2}}(\alpha -\phi )+...$ $\Rightarrow$$z=\frac{1}{1-\cos (\theta +\phi )\cos (\theta -\phi )}$ $\Rightarrow$$z=\frac{1}{1-\left( \frac{1}{x}-\frac{1}{y} \right)}=\frac{xy}{xy-(y-x)}$ $\Rightarrow$$xyz-(zy-zx)=xy$$\Rightarrow$$xyz-xy=yz-xz$