Question

If$x^{2}y=y-2x$ and $x^2<1$. then $a\times [x+\frac{1}{3}{x}^3+\frac{1}{5}{x}^5+.....]=y+\frac{1}{3}{y}^3+\frac{1}{5}{y}^5+......$ find $a$

Answer 1

Given,

$x^{2}y=y-2x$

or, $(-x^2+1)y=2x$

or, $y=\frac{2x}{1-x^2}$.......(1)

Also given, $a\times [x+\frac{1}{3}{x}^3+\frac{1}{5}{x}^5+.....]=y+\frac{1}{3}{y}^3+\frac{1}{5}{y}^5+......$

or, $a{\tan }^{-1}x={\tan }^{-1}y$ [ Using the series for ${\tan }^{-1}x$ and ${\tan }^{-1}y$]

or, $a{\tan }^{-1}x={\tan }^{-1}\frac{2x}{1-x^2}$

or, $a{\tan }^{-1}x={\tan }^{-1}\frac{2x}{1-x^2}$ [Using (1)]

or, $a{\tan }^{-1}x=2{\tan }^{-1}x$

or, $a=2$.