Question

$\text { If } \quad x=\cos 2 \theta-2 \cos ^2 2 \theta+3 \cos ^3 2 \theta-4 \cos ^4 2 \theta+\ldots \ldots \infty $
$\text { and } y=\cos 2 \theta+2 \cos ^2 2 \theta+3 \cos ^3 2 \theta+4 \cos ^4 2 \theta+\ldots \ldots \infty $
$\text { where } \theta \in\left(0, \frac{\pi}{4}\right) \text {, then find least integral value of }\left(\frac{1}{x}+\frac{1}{y}\right)$ .

Answer 1

$x=\cos 2 \theta-2 \cos ^2 2 \theta+3 \cos ^3 2 \theta \ldots \ldots \infty$
$\frac{-x \cos 2 \theta=-\cos ^2 2 \theta+2 \cos ^3 2 \theta+\ldots \ldots \infty}{x(1+\cos 2 \theta)=\cos 2 \theta-\cos ^2 2 \theta+\cos ^3 2 \theta \ldots \ldots \infty}$
$=\frac{\cos 2 \theta}{1+\cos 2 \theta}$
$\therefore x =\frac{\cos 2 \theta}{(1+\cos 2 \theta)^2} $
$\text { Similarly, } y =\frac{\cos 2 \theta}{(1-\cos 2 \theta)^2}$
$\therefore \frac{1}{x}+\frac{1}{y}=\frac{(1+\cos 2 \theta)^2+(1-\cos 2 \theta)^2}{\cos 2 \theta}=\frac{2\left(1+\cos ^2 2 \theta\right)}{\cos 2 \theta}=2(\sec 2 \theta+\cos 2 \theta)$
$\frac{1}{x}+\frac{1}{y} >4$
$\therefore $ least integral value of $\frac{1}{ x }+\frac{1}{ y }$ is 5