The ratio of the coefficient of $x^{10}\mathrm{in}{\left(1-x^2\right)}^{10}$and the term independent of x $\mathrm{in} {\left(\mathrm{x}-\frac{2}{\mathrm{x}}\right)}^{10}$,is

 Given, ${\left(1-x^2\right)}^{10}\text{ and }{\left(x-\frac{2}{x}\right)}^{10}$

To find Ratio of coefficient of $x^{10}\text{ in }{\left(1-x^2\right)}^{10}$ and the term of independent of $\mathrm{xin}{\left(\mathrm{x}-\frac{2}{\mathrm{x}}\right)}^{10}$

Now, let $(\mathrm{r}+1) \mathrm{th}$term in ${\left(1-{\mathrm{x}}^2\right)}^{10}$contains ${\mathrm{x}}^{10}$

Thus, coefficient of ${\mathrm{x}}^{10}\text{ is }{-}^{10}{\mathrm{C}}_5$

Again, let term be independent of x in ${\left(\mathrm{x}-\frac{2}{\mathrm{x}}\right)}^{10}$

$\begin{array}{r}\therefore {\mathrm{T}}_{\mathrm{r}+1}{=}^{10}{\mathrm{C}}_{\mathrm{r}}(\mathrm{x}{)}^{10-\mathrm{r}}{\left(-\frac{2}{\mathrm{x}}\right)}^{\mathrm{r}}\\ ={(-1{)}^{\mathrm{r}}}^{10}{\mathrm{C}}_{\mathrm{r}}\left(2{)}^{\mathrm{r}}\right(\mathrm{x}{)}^{10-2\mathrm{r}}\end{array}$

To get term independent of x, put 10 - 2r =0 =r r = b

Thus, the term independent of X in ${\left(\mathrm{x}-\frac{2}{\mathrm{x}}\right)}^{10}\text{ is }{-}^{10}{\mathrm{C}}_5(2{)}^5$

.'. Required ratio $=\frac{{-}^{10}{\mathrm{C}}_5}{{-}^{10}{\mathrm{C}}_5(2{)}^5}=\frac{1}{(2{)}^5}=\frac{1}{32}=1:32$