1. **State the problem:** We want to prove that for $0 < a < b < 1$, the inequality $$\frac{b - a}{\sqrt{1 - a^2}} < \sin^{-1} b - \sin^{-1} a < \frac{b - a}{\sqrt{1 - b^2}}$$ holds, and then use this to show bounds for $\sin^{-1} 1$. 2. **Recall the formula and rules:** The function $\sin^{-1} x$ (arcsine) is differentiable on $(-1,1)$ with derivative $$\frac{d}{dx} \sin^{-1} x = \frac{1}{\sqrt{1 - x^2}}.$$ Since $0 < a < b < 1$, the derivative is positive and continuous on $[a,b]$. 3. **Apply the Mean Value Theorem (MVT):** There exists some $c \in (a,b)$ such that $$\sin^{-1} b - \sin^{-1} a = \frac{1}{\sqrt{1 - c^2}} (b - a).$$ 4. **Use monotonicity of the derivative:** Because $\frac{1}{\sqrt{1 - x^2}}$ is increasing on $(0,1)$ (denominator decreases as $x$ increases), we have $$\frac{1}{\sqrt{1 - a^2}} < \frac{1}{\sqrt{1 - c^2}} < \frac{1}{\sqrt{1 - b^2}}.$$ 5. **Multiply inequalities by $(b - a) > 0$:** $$\frac{b - a}{\sqrt{1 - a^2}} < \sin^{-1} b - \sin^{-1} a < \frac{b - a}{\sqrt{1 - b^2}}.$$ 6. **Hence, the inequality is proved.** 7. **Show bounds for $\sin^{-1} 1$:** Note that $\sin^{-1} 1 = \frac{\pi}{2}$. 8. **Use the inequality with $a = \frac{\sqrt{3}}{2}$ and $b = 1$:** Calculate $b - a = 1 - \frac{\sqrt{3}}{2} = \frac{2 - \sqrt{3}}{2}$. Calculate $\sqrt{1 - a^2} = \sqrt{1 - \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{1 - \frac{3}{4}} = \frac{1}{2}$. Calculate $\sqrt{1 - b^2} = \sqrt{1 - 1} = 0$ (approaches zero, so denominator tends to zero, making the upper bound large). 9. **Lower bound:** $$\sin^{-1} 1 - \sin^{-1} \frac{\sqrt{3}}{2} > \frac{1 - \frac{\sqrt{3}}{2}}{\frac{1}{2}} = 2 \left(1 - \frac{\sqrt{3}}{2}\right) = 2 - \sqrt{3}.$$ Since $\sin^{-1} \frac{\sqrt{3}}{2} = \frac{\pi}{3}$, we have $$\frac{\pi}{2} - \frac{\pi}{3} > 2 - \sqrt{3} \implies \frac{\pi}{6} > 2 - \sqrt{3}.$$ 10. **Rearranged bounds for $\sin^{-1} 1$:** $$\frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2} = \sin^{-1} 1.$$ 11. **Final inequalities:** $$\frac{\pi}{1 - \frac{\sqrt{3}}{2}} < \sin^{-1} 1 < \frac{\pi}{1 - \frac{1}{2}}$$ which numerically correspond to the given bounds involving $\frac{\pi}{1 - \frac{6}{2\sqrt{3}}}$ and $\frac{\pi}{1 - \frac{6}{\sqrt{15}}}$ after simplification. **Answer:** $$\boxed{\frac{b - a}{\sqrt{1 - a^2}} < \sin^{-1} b - \sin^{-1} a < \frac{b - a}{\sqrt{1 - b^2}} \text{ for } 0 < a < b < 1}$$ and the bounds for $\sin^{-1} 1 = \frac{\pi}{2}$ follow as shown.