Question

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

Answer 1

$\Rightarrow {\cot }^{-1}\left(\frac{ab+1}{a-b}\right)+{\cot }^{-1}\left(\frac{bc+1}{b-c}\right)+{\cot }^{-1}\left(\frac{ca+1}{c-a}\right)$

$\Rightarrow {\tan }^{-1}\left(\frac{a-b}{ab+1}\right)+{\tan }^{-1}\left(\frac{b-c}{bc+1}\right)+{\tan }^{-1}\left(\frac{c-a}{ca+1}\right)$

$\Rightarrow {\tan }^{-1}\left(a\right)-{\tan }^{-1}\left(b\right)+\tan b-{\tan }^{-1}c+{\tan }^{-1}c-{\tan }^{-1}a$

$\Rightarrow 0$