If m and M are the least and greatest values of ${\left({\cos }^{-1}\mathrm{x}\right)}^2+{\left({\sin }^{-1}\mathrm{x}\right)}^2, \mathrm{then} \frac{\mathrm{M}}{\mathrm{m}}=$

$$\begin{gathered} let f\left(x\right)={\left({\cos }^{-1}\mathrm{x}\right)}^2+{\left({\sin }^{-1}\mathrm{x}\right)}^2={\left({\cos }^{-1}\mathrm{x}+{\sin }^{-1}\mathrm{x}\right)}^2-2{\sin }^{-1}{\mathrm{xcos}}^{-1}\mathrm{x} \\ =\frac{{\mathrm{\pi }}^2}{4}-2{\sin }^{-1}\mathrm{x}\left(\frac{\mathrm{\pi }}{2}-{\sin }^{-1}\mathrm{x}\right)=\frac{{\mathrm{\pi }}^2}{4}-2{\sin }^{-1}\mathrm{x}\left(\frac{\mathrm{\pi }}{2}-{\sin }^{-1}\mathrm{x}\right)\to \left(1\right) \\ \mathrm{put} \mathrm{a}={\sin }^{-1}\mathrm{x} \mathrm{then} \left(1\right) \mathrm{becomes} \\ \mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{\pi }}^2}{4}-\mathrm{\pi a}+2{\mathrm{a}}^2 \\ =2({\mathrm{a}}^2-\frac{\mathrm{\pi }}{2}\mathrm{a})+\frac{{\mathrm{\pi }}^2}{4} \\ =2({\mathrm{a}}^2-2(\mathrm{a}\left)\right(\frac{\mathrm{\pi }}{4})+\frac{{\mathrm{\pi }}^2}{16})-\frac{{\mathrm{\pi }}^2}{8}+\frac{{\mathrm{\pi }}^2}{4} \\ =2{(\mathrm{a}-\frac{\mathrm{\pi }}{4})}^2+\frac{{\mathrm{\pi }}^2}{8} \\ \mathrm{sub} \mathrm{a}=\frac{\mathrm{\pi }}{4} \mathrm{then} \mathrm{function} \mathrm{having} \mathrm{minimum} \mathrm{value} \\ \mathrm{f}\left(\frac{\mathrm{\pi }}{4}\right)=\frac{{\mathrm{\pi }}^2}{8} \\ \therefore \mathrm{m}=\frac{{\mathrm{\pi }}^2}{8} \\ sub a=\frac{-\pi }{2} then function having maximum value \\ f\left(\frac{-\pi }{2}\right)=\frac{5{\pi }^2}{4} \\ \mathrm{f}(-1)=\frac{5{\mathrm{\pi }}^2}{4} \\ \therefore M=\frac{5{\mathrm{\pi }}^2}{4} \\ \therefore \frac{\mathrm{M}}{\mathrm{m}}=10 \end{gathered}$$