Question

Prove that: ${\tan }^{-1}\left(\frac{a-b}{1+ab}\right)+{\tan }^{-1}\left(\frac{b-c}{1+bc}\right)+{\tan }^{-1}\left(\frac{c-a}{1+ca}\right)=0$.

Answer 1

Let us assume that $a=\tan \alpha$, $b=\tan \beta$ and $c=\tan \gamma$

$\therefore {\tan }^{-1}\left(\frac{a-b}{1+ab}\right)={\tan }^{-1}\left(\frac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta }\right)$

$={\tan }^{-1}\left(\tan \right(\alpha -\beta \left)\right)$

$=\alpha -\beta$

Similarly, ${\tan }^{-1}\left(\frac{b-c}{1+bc}\right)=\beta -\gamma$ and ${\tan }^{-1}\left(\frac{c-a}{1+ac}\right)=\gamma -\alpha$

$\therefore {\tan }^{-1}\left(\frac{a-b}{1+ab}\right)+{\tan }^{-1}\left(\frac{b-c}{1+bc}\right)+{\tan }^{-1}\left(\frac{c-a}{1+ac}\right)=\alpha -\beta +\beta -\gamma +\gamma -\alpha =0$