Verified Solution • 2025–26 Session moderate

QJEEM25Ma76b7: JEE Main 2025 Mathematics - Class 12 JEEM25Ma76b7 Solution: If , then the expression is equal to :

If \alpha>\beta>\gamma>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\} is equal to :

Multiple Choice Options

  • A 3π3 \pi
  • B π2−(α+β+γ)\frac{\pi}{2}-(\alpha+\beta+\gamma)
  • C π\pi
  • D 0

Step-by-Step Verified Solution

amp;⇒cot⁡−1(αβ+1α−β)+cot⁡−1(βγ+1β−γ)+cot⁡−1(αγ+1γ−α)amp;⇒tan⁡−1(α−β1+αβ)+tan⁡−1(β−γ1+βγ)+π+tan⁡−1(γ−α1+γα)amp;⇒(tan⁡−1α−tan⁡−1β)+(tan⁡−1β−tan⁡−1γ)+(π+tan⁡−1γ−tan⁡−1α)amp;⇒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}

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