If $\alpha\beta\gamma0$, then the expression $\cot^{-1}\left\{\beta+\frac{(1+\beta^2)}{(\alpha-\beta)}\right\}+\cot^{-1}\left\{\gamma+\frac{(1+\gamma^2)}{(\beta-\gamma)}\right\}+\cot^{-1}\left\{\alpha+\frac{(1+\alpha^2)}{(\gamma-\alpha)}\right\}$ is equal to :

Question

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

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] $\beta+\frac{1+\beta^2}{\alpha-\beta}=\frac{1+\alpha\beta}{\alpha-\beta}$, so the first term is $	an^{-1}\frac{\alpha-\beta}{1+\alpha\beta}=	an^{-1}\alpha-	an^{-1}\beta$. Similarly the second is $	an^{-1}\beta-	an^{-1}\gamma$. The third has a negative argument $\frac{1+\alpha\gamma}{\gamma-\alpha}$, so it equals $\pi+	an^{-1}\gamma-	an^{-1}\alpha$. The sum is $\pi$. Answer (1).

Q. If $\alpha>\beta>\gamma>0$, then the expression $\cot^{-1}\left\{\beta+\frac{(1+\beta^2)}{(\alpha-\beta)}\right\}+\cot^{-1}\left\{\gamma+\frac{(1+\gamma^2)}{(\beta-\gamma)}\right\}+\cot^{-1}\left\{\alpha+\frac{(1+\alpha^2)}{(\gamma-\alpha)}\right\}$ is equal to :

Explanation

Reviewed by Dr. Abubacker

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