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Mathematics · Inverse Trigonometric Functions

JEE Main 2025 — 4 April, Evening Shift — Question 37

The sum of the infinite series cot⁡−1(74)+cot⁡−1(194)+cot⁡−1(394)+cot⁡−1674+…\cot ^{-1}\left(\frac{7}{4}\right)+\cot ^{-1}\left(\frac{19}{4}\right)+\cot ^{-1}\left(\frac{39}{4}\right)+\cot ^{-1} \frac{67}{4}+\ldots is:

  1. Option A:

    π2−cot⁡−1(12)\frac{\pi}{2}-\cot ^{-1}\left(\frac{1}{2}\right)

  2. Option B:

    π2−tan⁡−1(12)\frac{\pi}{2}-\tan ^{-1}\left(\frac{1}{2}\right)

    Correct
  3. Option C:

    π2+tan⁡−1(12)\frac{\pi}{2}+\tan ^{-1}\left(\frac{1}{2}\right)

  4. Option D:

    π2+cot⁡−1(12)\frac{\pi}{2}+\cot ^{-1}\left(\frac{1}{2}\right)

Answer: B

Step-by-step solution

cot⁡−1(74)+cot⁡−1(194)+cot⁡−1(394)+cot⁡−1(674)+…\cot ^{-1}\left(\frac{7}{4}\right)+\cot ^{-1}\left(\frac{19}{4}\right)+\cot ^{-1}\left(\frac{39}{4}\right)+\cot ^{-1}\left(\frac{67}{4}\right)+\ldots

Tr=cot⁡−1(4r2+34)Tr=tan⁡−1(1(34+r2))Tr=tan⁡−1((r+12)−(r−12)1+r2−1/4)Tr=tan⁡−1((r+12)−(r−12)1+(r+12)(r−12))Tr=tan⁡−1(r+12)−tan⁡−1(r−12)T1=tan⁡−1(32)−tan⁡−1(12)\begin{aligned} & T_{r}=\cot ^{-1}\left(\frac{4 r^{2}+3}{4}\right) \\& T_{r}=\tan ^{-1}\left(\frac{1}{\left(\frac{3}{4}+r^{2}\right)}\right) \\& T_{r}=\tan ^{-1}\left(\frac{\left(r+\frac{1}{2}\right)-\left(r-\frac{1}{2}\right)}{1+r^{2}-1 / 4}\right) \\& T_{r}=\tan ^{-1}\left(\frac{\left(r+\frac{1}{2}\right)-\left(r-\frac{1}{2}\right)}{1+\left(r+\frac{1}{2}\right)\left(r-\frac{1}{2}\right)}\right) \\& T_{r}=\tan ^{-1}\left(r+\frac{1}{2}\right)-\tan ^{-1}\left(r-\frac{1}{2}\right) \\& T_{1}=\tan ^{-1}\left(\frac{3}{2}\right)-\tan ^{-1}\left(\frac{1}{2}\right) \end{aligned} T2=tan⁡−1(52)−tan⁡−1(32)⋮⋮Tn=tan⁡−1(2n+12)−tan⁡−1(12)ΣTr=tan⁡−1(2n+12)−tan⁡−1(12)ΣTr=π2−tan⁡−1(12)\begin{aligned} & T_{2}=\tan ^{-1}\left(\frac{5}{2}\right)-\tan ^{-1}\left(\frac{3}{2}\right) \\& \vdots \\& \vdots \\& T_{n}=\tan ^{-1}\left(\frac{2 n+1}{2}\right)-\tan ^{-1}\left(\frac{1}{2}\right) \\& \Sigma T_{r}=\tan ^{-1}\left(\frac{2 n+1}{2}\right)-\tan ^{-1}\left(\frac{1}{2}\right) \\& \Sigma T_{r}=\frac{\pi}{2}-\tan ^{-1}\left(\frac{1}{2}\right) \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
4 April, Evening Shift
Subject
Mathematics
Chapter
Inverse Trigonometric Functions
Topic
Summation of Series involving ITFs