Question

If $x={\sum }_{n=0}^{\infty }{\cos }^{2n}\theta ,y={\sum }_{n=0}^{\infty }{\sin }^{2n}\theta$
$z={\sum }_{n=0}^{\infty }{\cos }^{2n}\theta {\sin }^{2n}\theta$ and $0<\theta <\frac{\pi }{2}$, then

• A. $xz+yz=xy+z$
• B. $xyz=yz+x$
• C. $xy+z=xy+zx$
• D. $x+y+z=xyz+z$

Answer 1

Answer: (A)

$\because x=\sum _{n=0}^{\infty }{\cos }^{2n}\theta =\frac{1}{1-{\cos }^2\theta }$
$=\frac{1}{{\sin }^2\theta }\left\{\because 0<\theta <\frac{\pi }{2}\right\}$
$y=\sum _{n=0}^{\infty }{\sin }^{2n}\theta =\frac{1}{1-{\sin }^2\theta }$
$=\frac{1}{{\cos }^2\theta }$
$z=\sum _{n=0}^{\infty }{\cos }^{2n}\theta {\sin }^{2n}\theta$
$=\frac{1}{1-{\cos }^2\theta {\sin }^2\theta }$
$\therefore z=\frac{1}{1-\frac{1}{xy}}$
$\Rightarrow xyz-z=xy...(i)$

$\because \frac{1}{x}+\frac{1}{y}={\sin }^2\theta +{\cos }^2\theta =1$
$\Rightarrow x+y=xy......(ii)$
From Eqs. (i) and (ii),
$(x+y)z-z=xy$
$\Rightarrow xz+yz=xy+z$