Question

If $0 < a < b < c$, then
$cot^{-1}\left(\frac{ab+1}{a-b}\right)+cot^{-1}\left(\frac{bc+1}{b-c}\right)+cot^{-1}\left(\frac{ca+1}{c-a}\right)=$

• A. $0$
• B. $\pi$
• C. $2\pi$
• D. None of these

Answer 1

Answer: (C)

$\because a-b < 0,$ so $cot^{-1} \frac{ab+1}{a-b}=cot^{-1}\,b-cot^{-1}\,a+\pi$
$b-c < 0,$ so $cot^{-1} \frac{bc+1}{b-c}=cot^{-1}\,c-cot^{-1}\,b+\pi$
$c-b < 0,$ so $cot^{-1} \frac{ca+1}{c-a}=cot^{-1}\,a-cot^{-1}\,c$
Adding we get
$cot^{-1}\left(\frac{ab+1}{a-b}\right)+cot^{-1}\left(\frac{bc+1}{b-c}\right)+cot^{-1}\left(\frac{ca+1}{c-a}\right)=2\pi$