The tautochrone property (meaning equal time) is one of the dynamic properties of an inverted cycloid. This means that if one puts two objects at different positions on a inverted cycloidial shaped slide, then both objects will reach the bottom of that slide at the exact same time. I do not know how to show this mathematically.
However, I do know the following:
The mathematical formula for the tautochrone property: $\tau=\pi\sqrt{\frac{a}{g}}$
I began with the following derivation to determine the time it takes to reach the bottom for any starting point along the curve:
$\tau=\sqrt{\frac{a}{g}}\int_{\theta_0}^\pi \mathrm{\sqrt{\frac{1-\cos{\theta}}{\cos{\theta_0}-\cos{\theta}}}\mathrm{d}\theta}$
I began simplifying the derivation with trigonometric half-angle properties:
$\sqrt{2}\sin{\frac{\theta}{2}}=\sqrt{1-\cos{\theta}}$
$\cos{\theta}=2\cos^2{\frac{\theta}{2}}-1$
The root 2's cancel out:
$\tau=\pi\sqrt{\frac{a}{g}}\int_{\theta_0}^\pi \mathrm{\frac{\sin{\frac{\theta}{2}}}{\sqrt{\cos^2{\frac{\theta_0}{2}}-\cos^2{\frac{\theta}{2}}}}\mathrm{d}\theta}$
How do I proceed from here?
Now let $u=\cos\frac{\theta}{2}$, so $du=-\frac{1}{2}\sin\frac{\theta}{2}d\theta$ to get
$\displaystyle 2\sqrt{\frac{a}{g}}\int_0^{\cos \theta_{0}/2}\frac{1}{\sqrt{\cos^2\frac{\theta_0}{2}-u^2}}du=2\sqrt{\frac{a}{g}}\left[\sin^{-1}\frac{u}{\cos\frac{\theta_0}{2}}\right]_0^{\cos\theta_0/2}=2\sqrt{\frac{a}{g}}\sin^{-1}1=\pi\sqrt{\frac{a}{g}}$.