If minimum value of ${\left({\sin }^{-1}x\right)}^2+{\left({\cos }^{-1}x\right)}^2$  is $\frac{{\pi }^2}{k}$then the value of K is

${\left({\sin }^{-1}x\right)}^2+{\left({\cos }^{-1}x\right)}^2=$${\left({\sin }^{-1}x\right)}^2+{(\frac{\pi }{2}-{\sin }^{-1}x)}^2$

$=2{\left({\sin }^{-1}x\right)}^2-\pi {\sin }^{-1}x+\frac{{\pi }^2}{4}$

 Its minimum value is $\frac{\mathrm{4.2.}\frac{{\pi }^2}{4}-{\pi }^2}{8}=\frac{{\pi }^2}{8}$