Question

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

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

Answer 1

Answer: (A)

$x+y+z$

$x=1+{\cos }^2\theta +{\cos }^4\theta +...\infty$
Hence sum of G.P is
$=\frac{1}{1-{\cos }^2\theta }$
$=\frac{1}{{\sin }^2\theta }$
$y=1+{\sin }^2\theta +{\sin }^4\theta +...\infty$
Hence sum of G.P is
$=\frac{1}{1-{\sin }^2\theta }$

$=\frac{1}{{\cos }^2\theta }$

$z=1+{\cos }^2\theta .{\sin }^2\theta +{\cos }^4\theta .{\sin }^4\theta ...\infty$

$=\frac{1}{1-{\cos }^2\theta .{\sin }^2\theta }$
Hence $x.y.z$
$=\frac{1}{{\cos }^2\theta .{\sin }^2\theta }.\frac{1}{1-{\cos }^2\theta .{\sin }^2\theta }$

$=\frac{1}{{\cos }^2\theta .{\sin }^2\theta (1-{\cos }^2\theta .{\sin }^2\theta )}$

$=\frac{1-{\cos }^2\theta .{\sin }^2\theta +{\cos }^2\theta .{\sin }^2\theta }{{\cos }^2\theta .{\sin }^2\theta (1-{\cos }^2\theta .{\sin }^2\theta )}$

$=\frac{1}{{\cos }^2\theta .{\sin }^2\theta }+\frac{1}{1-{\cos }^2\theta .{\sin }^2\theta }$

$=\frac{{\cos }^2\theta +{\sin }^2\theta }{{\cos }^2\theta .{\sin }^2\theta }+\frac{1}{1-{\cos }^2\theta .{\sin }^2\theta }$

$=\frac{1}{{\cos }^2\theta }+\frac{1}{si{n}^2\theta }+\frac{1}{1-{\cos }^2\theta .{\sin }^2\theta }$

$=x+y+z$