Question

If \( a^2 + b^2 + c^2 = r^2 \), then the value of \( \tan^{-1}\left(\frac{ab}{cr}\right) + \tan^{-1\left(\frac{bc}{ar}\right) + \tan^{-1}\left(\frac{ca}{br}\right) = \)}

• A. \( \frac{\pi}{2} \)
• B. \( \frac{\pi}{3} \)
• C. \( \frac{\pi}{4} \)
• D. \( \frac{\pi}{6} \)

Hint:

Identity: If \( xy+yz+zx = 1 \), then the sum of their inverse tangents is \( 90^\circ \).

Answer 1

The Correct Option is A

Step 1: Concept Use the formula for \( \tan^{-1} x + \tan^{-1} y + \tan^{-1} z \).

Step 2: Meaning Let \( x = \frac{ab}{cr}, y = \frac{bc}{ar}, z = \frac{ca}{br} \).

Step 3: Analysis Check the condition \( xy + yz + zx \). \( xy = \frac{b^2}{r^2} \), \( yz = \frac{c^2}{r^2} \), \( zx = \frac{a^2}{r^2} \). \( xy + yz + zx = \frac{a^2+b^2+c^2}{r^2} = \frac{r^2}{r^2} = 1 \).

Step 4: Conclusion If \( xy+yz+zx = 1 \), then \( \tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \pi/2 \). Final Answer: (A)