Question

lf $x=\sum _{n=0}^{\infty }{\sin }^{2n}\theta ,y=\sum _{n=0}^{\infty }{\cos }^{2n}\theta ,z=\sum _{n=0}^{\infty }{\sin }^{2n}\theta {\cos }^{2n}\theta$, then

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

Answer 1

Answer: (A), (D)

The correct options are
A $xyz=xy+z$
D $xyz=x+y+z$

$x=1+{\sin }^2\theta +{\sin }^4\theta +{\sin }^6\theta .....\infty$
GP with common ratio $={\sin }^2\theta$
$x=\frac{1}{1-{\sin }^2\theta }=\frac{1}{{\cos }^2\theta }$
$y=1+{\cos }^2\theta +{\cos }^4\theta +{\cos }^6\theta .....\infty$
GP with common ratio$={\cos }^2\theta$
$y=\frac{1}{1-{\cos }^2\theta }=\frac{1}{{\sin }^2\theta }$
$1+{\sin }^2\theta {\cos }^2\theta +{\sin }^4\theta {\cos }^4\theta ........\infty$
GP with common ratio$={\sin }^2\theta {\cos }^2\theta$
$z=\frac{1}{1-{{\sin }^2\theta \cos }^2\theta }=\frac{1}{1-\frac{1}{xy}}=\frac{xy}{xy-1}$
$xyz=xy+z$...........(1)
${\sin }^2\theta +{\cos }^2\theta =1$
$1=\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}$ ......(2)
$xy=y+x$...............(3)
Substitue value of (3) in (1)
$x+y+z=xyz$