Mathematics · 2025
JEE Main · 24 January 2025, Shift 2 · Q14
If α>β>γ>0, then the expression cot^-1{β+((1+β^2))/((α-β))}+ cot^-1{γ+((1+γ^2))/((β-γ))}+ cot^-1{α+((1+α^2))/((γ-α))} is equal to:
If $\displaystyle \alpha>\beta>\gamma>0$, then the expression
$\displaystyle \cot ^{-1}\left\{\beta+\frac{\left(1+\beta^2\right)}{(\alpha-\beta)}\right\}+\cot ^{-1}\left\{\gamma+\frac{\left(1+\gamma^2\right)}{(\beta-\gamma)}\right\}+\cot ^{-1}\left\{\alpha+\frac{\left(1+\alpha^2\right)}{(\gamma-\alpha)}\right\}$ is equal to :
Official answer
From NTA’s final answer key for this paper.
(4)
$\displaystyle \pi$
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JEE Main 2025 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.