Let (a , b) $\in$(0,2 $\pi$) be the largest interval for which ${\sin }^{-1}\left(\sin \theta \right)-{\cos }^{-1}\left(\sin \theta \right)>0, \theta \in (0,2\pi )$ holds. If  $\alpha x^2+\beta x+{\sin }^{-1}(x^2-8x+17)+{\cos }^{-1}(x^2-8x+17)=0$  and  $\alpha -\beta =b-a$ then $\alpha$ is equal to.

$$\begin{gathered} {\sin }^{-1}\left(\sin \theta \right)-{\cos }^{-1}\left(\sin \theta \right)>0 \\ {\sin }^{-1}\left(\sin \theta \right)-\frac{\pi }{2}+{\sin }^{-1}\left(\sin \theta \right)>0 \\ {\sin }^{-1}\left(\sin \theta \right)>\frac{\pi }{4} \\ \sin \theta >\frac{1}{\sqrt{2}}\Rightarrow \theta \in (\frac{\pi }{4},\frac{3\pi }{4})=(a,b) \\ a=\frac{\pi }{4},b=\frac{3\pi }{4} \\ b-a=\frac{\pi }{2}=\alpha -\beta \\ \beta =\alpha -\frac{\pi }{2}\to \left(1\right) \end{gathered}$$

$\alpha x^2+\beta x+{\sin }^{-1}({(x-4)}^2+1)+{\cos }^{-1}({(x-4)}^2+1)=0$ this is possible only if  $x=4$

$$\begin{gathered} 16\alpha +4\beta +\frac{\pi }{2}+0=0 \\ 16\alpha +4\beta +\frac{\pi }{2}=0 \\ 16\alpha +4(\alpha -\frac{\pi }{2})+\frac{\pi }{2}=0 \\ 20\alpha -2\pi +\frac{\pi }{2}=0 \\ 20\alpha =\frac{3\pi }{2} \\ \alpha =\frac{3\pi }{40} \end{gathered}$$