Let (a, b) subset (0, 2pi) be the largest int | vedclass

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(D) Given $\sin^{-1}(\sin 	heta) - \cos^{-1}(\sin 	heta) > 0$.Using $\cos^{-1}(x) = \frac{\pi}{2} - \sin^{-1}(x)$,we get $\sin^{-1}(\sin 	heta) - (

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$\frac{\pi}{12}$

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(D) Given $\sin^{-1}(\sin 	heta) - \cos^{-1}(\sin 	heta) > 0$.Using $\cos^{-1}(x) = \frac{\pi}{2} - \sin^{-1}(x)$,we get $\sin^{-1}(\sin 	heta) - (\frac{\pi}{2} - \sin^{-1}(\sin 	heta)) > 0$.$2 \sin^{-1}(\sin 	heta) > \frac{\pi}{2} \Rightarrow \sin^{-1}(\sin 	heta) > \frac{\pi}{4}$.This implies $\sin 	heta > \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.For $	heta \in (0, 2\pi)$,$\sin 	heta > \frac{1}{\sqrt{2}}$ holds for $	heta \in (\frac{\pi}{4}, \frac{3\pi}{4})$.Thus,$(a, b) = (\frac{\pi}{4}, \frac{3\pi}{4})$,so $b - a = \frac{3\pi}{4} - \frac{\pi}{4} = \frac{\pi}{2}$.Given $\alpha - \beta = b - a = \frac{\pi}{2}$,so $\beta = \alpha - \frac{\pi}{2}$.The equation is $\alpha x^2 + \beta x + \sin^{-1}((x-3)^2 + 1) + \cos^{-1}((x-3)^2 + 1) = 0$.Since $\sin^{-1}(y) + \cos^{-1}(y) = \frac{\pi}{2}$ for $y \in [-1, 1]$,and here $y = (x-3)^2 + 1$,we must have $(x-3)^2 + 1 \in [-1, 1]$.Since $(x-3)^2 \ge 0$,$(x-3)^2 + 1 \ge 1$. Thus,we must have $(x-3)^2 + 1 = 1$,which implies $x = 3$.Substituting $x = 3$ into the equation: $\alpha(3)^2 + \beta(3) + \frac{\pi}{2} = 0$.$9\alpha + 3(\alpha - \frac{\pi}{2}) + \frac{\pi}{2} = 0$.$9\alpha + 3\alpha - \frac{3\pi}{2} + \frac{\pi}{2} = 0$.$12\alpha - \pi = 0 \Rightarrow \alpha = \frac{\pi}{12}$.

Let $(a, b) \subset (0, 2\pi)$ be the largest interval for which $\sin^{-1}(\sin \theta) - \cos^{-1}(\sin \theta) > 0$ holds for $\theta \in (0, 2\pi)$. If $\alpha x^2 + \beta x + \sin^{-1}(x^2 - 6x + 10) + \cos^{-1}(x^2 - 6x + 10) = 0$ and $\alpha - \beta = b - a$,then $\alpha$ is equal to:

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