JEE Main 2021Inverse Trigonometric FunctionsInverse Trigonometric FunctionsMediumQuestionIf α>β>γ>0\alpha>\beta>\gamma>0α>β>γ>0, then the expression cot−1{β+(1+β2)(α−β)}+cot−1{γ+(1+γ2)(β−γ)}+cot−1{α+(1+α2)(γ−α)}\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\}cot−1{β+(α−β)(1+β2)}+cot−1{γ+(β−γ)(1+γ2)}+cot−1{α+(γ−α)(1+α2)} is equal to :OptionsA3π3 \pi3πBπ2−(α+β+γ)\frac{\pi}{2}-(\alpha+\beta+\gamma)2π−(α+β+γ)Cπ\piπD0Check AnswerHide SolutionSolution⇒cot−1(αβ+1α−β)+cot−1(βγ+1β−γ)+cot−1(αγ+1γ−α)⇒tan−1(α−β1+αβ)+tan−1(β−γ1+βγ)+π+tan−1(γ−α1+γα)⇒(tan−1α−tan−1β)+(tan−1β−tan−1γ)+(π+tan−1γ−tan−1α)⇒π\begin{aligned} & \Rightarrow \cot ^{-1}\left(\frac{\alpha \beta+1}{\alpha-\beta}\right)+\cot ^{-1}\left(\frac{\beta \gamma+1}{\beta-\gamma}\right)+\cot ^{-1}\left(\frac{\alpha \gamma+1}{\gamma-\alpha}\right) \\ & \Rightarrow \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) \\ & \Rightarrow\left(\tan ^{-1} \alpha-\tan ^{-1} \beta\right)+\left(\tan ^{-1} \beta-\tan ^{-1} \gamma\right)+\left(\pi+\tan ^{-1} \gamma-\tan ^{-1} \alpha\right) \\ & \Rightarrow \pi \end{aligned}⇒cot−1(α−βαβ+1)+cot−1(β−γβγ+1)+cot−1(γ−ααγ+1)⇒tan−1(1+αβα−β)+tan−1(1+βγβ−γ)+π+tan−1(1+γαγ−α)⇒(tan−1α−tan−1β)+(tan−1β−tan−1γ)+(π+tan−1γ−tan−1α)⇒π