If 0, then ^-1+1+^2-+^-1+1+^2-+^-1+1+^2- is equal to :

Question

If $\alpha>\beta>\gamma>0$, then $\cot^{-1}\left\{\beta+\dfrac{1+\beta^2}{\alpha-\beta}\right\}+\cot^{-1}\left\{\gamma+\dfrac{1+\gamma^2}{\beta-\gamma}\right\}+\cot^{-1}\left\{\alpha+\dfrac{1+\alpha^2}{\gamma-\alpha}\right\}$ is equal to :

Accepted Answer

Rewrite each term as $\cot^{-1}\left(\dfrac{\alpha\beta+1}{\alpha-\beta}\right)$, etc. Using $\cot^{-1}\left(\dfrac{xy+1}{x-y}\right)=\tan^{-1}\left(\dfrac{x-y}{1+xy}\right)$ for the first two brackets (positive denominators) and noting the third bracket's argument is negative (needing a $+\pi$ correction):

$$\tan^{-1}\left(\frac{\alpha-\beta}{1+\alpha\beta}\right)+\tan^{-1}\left(\frac{\beta-\gamma}{1+\beta\gamma}\right)+\pi+\tan^{-1}\left(\frac{\gamma-\alpha}{1+\gamma\alpha}\right)$$

$$=(\tan^{-1}\alpha-\tan^{-1}\beta)+(\tan^{-1}\beta-\tan^{-1}\gamma)+\pi+(\tan^{-1}\gamma-\tan^{-1}\alpha)=\pi.$$

Solutions · 1