The greatest and least values of (sin^-1 x)² + (cos^-1 x)² are a,b respectively

$\begin{array}{l}\mathrm{We} \mathrm{have}, {\left({\sin }^{-1}\mathrm{x}\right)}^2+{\left({\cos }^{-1}\mathrm{x}\right)}^2\\ ={\left({\sin }^{-1}\mathrm{x}+{\cos }^{-1}\mathrm{x}\right)}^2-2{\sin }^{-1}\mathrm{x}\cdot {\cos }^{-1}\mathrm{x}\\ =\frac{{\mathrm{\pi }}^2}{4}-2{\sin }^{-1}\mathrm{x}\left(\frac{\mathrm{\pi }}{2}-{\sin }^{-1}\mathrm{x}\right)\\ =\frac{{\mathrm{\pi }}^2}{4}-{\mathrm{\pi sin}}^{-1}\mathrm{x}+2{\left({\sin }^{-1}\mathrm{x}\right)}^2\\ =2\left[{\left({\sin }^{-1}\mathrm{x}\right)}^2-\frac{\mathrm{\pi }}{2}{\sin }^{-1}\mathrm{x}+\frac{{\mathrm{\pi }}^2}{8}\right]\\ =2\left[{\left({\sin }^{-1}\mathrm{x}-\frac{\mathrm{\pi }}{4}\right)}^2+\frac{{\mathrm{\pi }}^2}{16}\right]\end{array}$
Thus, the least value is, $\left(\frac{{\mathrm{\pi }}^2}{16}\right) \mathrm{i}.\mathrm{e. }\frac{{\mathrm{\pi }}^2}{8}$ and the greatest value is 
$2\left[{\left(\frac{-\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{4}\right)}^2+\frac{{\mathrm{\pi }}^2}{16}\right] \mathrm{i}.\mathrm{e}. \frac{5{\mathrm{\pi }}^2}{4}$