If $a^2+b^2+c^2=r^2$, then the value of $\tan ^{-1}\left(\frac{a b}{c r}\right)+\tan ^{-1}\left(\frac{b c}{a r}\right)+\tan ^{-1}\left(\frac{c a}{b r}\right)=$

Given the expression $S = \tan^{-1}\left(\frac{a b}{c r}\right) + \tan^{-1}\left(\frac{b c}{a r}\right) + \tan^{-1}\left(\frac{c a}{b r}\right)$, define $x = \frac{a b}{c r}$, $y = \frac{b c}{a r}$, and $z = \frac{c a}{b r}$.

The products $xy$, $yz$, and $zx$ simplify as: $xy = \frac{b^2}{r^2}$, $yz = \frac{c^2}{r^2}$, and $zx = \frac{a^2}{r^2}$. Their sum is $xy + yz + zx = \frac{a^2 + b^2 + c^2}{r^2}$.

Since $a^2 + b^2 + c^2 = r^2$, it follows that $xy + yz + zx = 1$. Given $x, y, z > 0$ for positive $a, b, c, r$, the identity $\tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \frac{\pi}{2}$ holds.

$S = \frac{\pi}{2}$