If $\alpha\beta\gamma0$, then the expression ${\cot }^{-1}\left{\beta +\frac{\left(1+{\beta }^{2}\right)}{(\alpha -\beta )}\right}$$+{\cot }^{-1}\left{\gamma +\frac{\left(1+{\gamma }^{2}\right)}{(\beta -\gamma )}\right}$$+{\cot }^{-1}\left{\alpha +\frac{\left(1+{\alpha }^{2}\right)}{(\gamma -\alpha )}\right}$ is equal to : [JEE Main 2025, 24 Jan (Shift 2)]

Question

If $\alpha>\beta>\gamma>0$, then the expression ${\cot }^{-1}\left{\beta +\frac{\left(1+{\beta }^{2}\right)}{(\alpha -\beta )}\right}$$+{\cot }^{-1}\left{\gamma +\frac{\left(1+{\gamma }^{2}\right)}{(\beta -\gamma )}\right}$$+{\cot }^{-1}\left{\alpha +\frac{\left(1+{\alpha }^{2}\right)}{(\gamma -\alpha )}\right}$ is equal to : [JEE Main 2025, 24 Jan (Shift 2)]

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

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If \(\alpha&gt;\beta&gt;\gamma&gt;0\), then the expression \({\cot }^{-1}\left{\beta +\frac{\left(1+{\beta }^{2}\right)}…

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