MHT CET · Maths · Inverse Trigonometric Functions
If \(a^2+b^2+c^2=r^2\), then the value of \(\tan ^{-1}\left(\frac{a b}{c r}\right)+\tan ^{-1}\left(\frac{b c}{a r}\right)+\tan ^{-1}\left(\frac{c a}{b r}\right)=\)
- A \(\frac{\pi}{2}\)
- B \(\frac{\pi}{3}\)
- C \(\frac{\pi}{4}\)
- D \(\frac{\pi}{6}\)
Answer & Solution
Correct Answer
(A) \(\frac{\pi}{2}\)
Step-by-step Solution
Detailed explanation
Let \(x=\frac{ab}{cr}, y=\frac{bc}{ar}, z=\frac{ca}{br}\). \(xy = \left(\frac{ab}{cr}\right)\left(\frac{bc}{ar}\right) = \frac{b^2}{r^2}\)
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