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Mathematics · Inverse Trigonometric Functions

JEE Main 2025 — 24 January, Evening Shift — Question 3

If α>β>γ>0\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:

  1. Option A:

    π2−(α+β+γ)\frac{\pi}{2}-(\alpha+\beta+\gamma)

  2. Option B:

    3π3 \pi

  3. Option C:

    0

  4. Option D:

    π\pi

    Correct

Answer: D

Step-by-step solution

⇒cot⁡−1(αβ+1α−β)+cot⁡−1(βγ+1β−γ)+cot⁡−1(αγ+1γ−α)\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)

⇒tan⁡−1(α−β1+αβ)+tan⁡−1(β−γ1+βγ)+π+tan⁡−1(γ−α1+γα)\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)

⇒(tan⁡−1α−tan⁡−1β)+(tan⁡−1β−tan⁡−1γ)+(π+tan⁡−1γ−tan⁡−1α)\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)

⇒π\Rightarrow \pi

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
24 January, Evening Shift
Subject
Mathematics
Chapter
Inverse Trigonometric Functions
Topic
Properties related to Inverse Trigonometric Functions
If α β γ 0 , then the expression cot -1 \ β+frac (1+β 2 ) (α-β) \… | JEE Main 2025 PYQ with Solution · DhiX AI