JEE Main 2025January, Unspecified (session-level dataset)MathematicsMediumTrigonometry118× since 2002If α>β>γ>0\alpha > \beta > \gamma > 0α>β>γ>0, then the expression cot−1{β+(1+β5)(α−β)}+cot−1{γ+(1+γ2)(β−γ)}+cot−1{α+(1+α2)(γ−α)}\cot^{-1} \left\{\beta + \frac{(1 + \beta^5)}{(\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\}cot−1{β+(α−β)(1+β5)}+cot−1{γ+(β−γ)(1+γ2)}+cot−1{α+(γ−α)(1+α2)} is equal to:AOption A. π\piπBOption B. 0COption C. π2−(α+β+γ)\frac{\pi}{2} - (\alpha + \beta + \gamma)2π−(α+β+γ)DOption D. 3π3 \pi3πCheckStep-by-step hintSkip · show solution
