Step 1: Understanding the Concept:
We need to evaluate the sum of three inverse tangent terms.
Let the terms be \( x = \frac{ab}{cr} \), \( y = \frac{bc}{ar} \), and \( z = \frac{ca}{br} \).
We know the identity for \( \tan^{-1} x + \tan^{-1} y + \tan^{-1} z \). The value heavily depends on the product pairs \( xy+yz+zx \).
Step 2: Key Formula or Approach:
Let \( S = \tan^{-1} x + \tan^{-1} y + \tan^{-1} z \).
Then \( \tan S = \frac{x + y + z - xyz}{1 - (xy + yz + zx)} \).
If \( xy + yz + zx = 1 \), the denominator becomes zero, which means \( \tan S \) is undefined, implying \( S = \frac{\pi}{2} \) (assuming positive principal values).
Step 3: Detailed Explanation:
Let's calculate the pairwise products \( xy, yz, \) and \( zx \):
\[ xy = \left(\frac{ab}{cr}\right) \left(\frac{bc}{ar}\right) = \frac{ab^2c}{acr^2} = \frac{b^2}{r^2} \]
\[ yz = \left(\frac{bc}{ar}\right) \left(\frac{ca}{br}\right) = \frac{bc^2a}{bar^2} = \frac{c^2}{r^2} \]
\[ zx = \left(\frac{ca}{br}\right) \left(\frac{ab}{cr}\right) = \frac{ca^2b}{cbr^2} = \frac{a^2}{r^2} \]
Now, find their sum:
\[ xy + yz + zx = \frac{b^2}{r^2} + \frac{c^2}{r^2} + \frac{a^2}{r^2} = \frac{a^2 + b^2 + c^2}{r^2} \]
We are given that \( a^2 + b^2 + c^2 = r^2 \). Substitute this into the sum:
\[ xy + yz + zx = \frac{r^2}{r^2} = 1 \]
Because \( xy + yz + zx = 1 \), the denominator of the \( \tan(A+B+C) \) formula becomes zero.
Thus, the sum of the angles is \( \frac{\pi}{2} \) (assuming \( a, b, c, r \) are positive, which puts the angles in the first quadrant).
Step 4: Final Answer:
The value of the expression is \( \frac{\pi}{2} \).