Statement – I  - If x = sin^-1(sin10) and y = cos^-1(cos10) then

 y-x = π
Statement – II  -The no. of solutions of the equation ${\sin }^{-1}[x^2+\frac{1}{3}]+{\cos }^{-1}[x^2-\frac{2}{3}]=x^2,x\in [-1,1]$  

and [.] denoted the greatest integer less than and equal to x is 0
 

$$\begin{gathered} x={\sin }^{-1}\left(\sin 10\right)={\sin }^{-1}\sin \left(3\pi -10\right)=3\pi -10 \\ \begin{array}{r}y={\cos }^{-1}(\cos 10)={\cos }^{-1}\cos (4\pi -10)=4\pi -10\\ y-x=\pi \end{array} \\ \begin{array}{r}\text{ II. }{\sin }^{-1}\left[x^2+\frac{1}{3}\right]+{\cos }^{-1}\left[x^2+\frac{1}{3}-1\right]=x^2\\ {\sin }^{-1}\left[x^2+\frac{1}{3}\right]+{\cos }^{-1}\left(\left[x^2+\frac{1}{3}\right]-1\right)=x^2\\ -1\le \left[x^2+\frac{1}{3}\right]\le 1:-1\le \left[x^2+\frac{1}{3}\right]-1\le 1\\ -1\le \left[x^2+\frac{1}{3}\right]\le 1:0\le \left[x^2+\frac{1}{3}\right]\le 2\\ \therefore 0\le \left[x^2+\frac{1}{3}\right]\le 1\\ \left[x^2+\frac{1}{3}\right]=0,1\end{array} \\ \text{ For }\left[x^2+\frac{1}{3}\right]=0,{\sin }^{-1}0+{\cos }^{-1}\left(-1\right)=x^2\Rightarrow x^2=\pi \text{ which are not possible } \\ \text{ For }\left|x^2+\frac{1}{3}\right|=1,{\sin }^{-1}1+{\cos }^{-1}\left(0\right)=x^2\Rightarrow x^2=\pi \text{ which are not possible } \\ \text{ Hence it has no solution } \end{gathered}$$