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Telescoping sum of tan^-1 or cot^-1 terms

This type of question, asked 7 times: 9 May 2024 S1, 14 May 2024 S2, 19 Apr 2025 S1, 26 Apr 2025 S2, 18 Apr 2026 S1, 20 Apr 2026 S1, 14 May 2026 S2. All 7 questions below, oldest first.

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  1. Question 1

    9 May 2024 S1

    113
    The value of cot⁡∑n=123cot⁡−1 ⁣(1+∑k=1n2k)\cot\sum_{n=1}^{23}\cot^{-1}\!\left(1+\sum_{k=1}^{n}2k\right) is

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  2. Question 2

    14 May 2024 S2

    134
    If ∑r=150tan⁡−112r2=p\sum_{r=1}^{50} \tan^{-1}\frac{1}{2r^2} = p, then tan⁡p\tan p is

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  3. Question 3

    19 Apr 2025 S1

    121
    The sum to infinite terms of the series tan⁡−1(13)+tan⁡−1(29)+……+tan⁡−1(2n−11+22n−1)+……\tan^{- 1}\left( \frac{1}{3} \right)+\tan^{- 1}\left( \frac{2}{9} \right)+ \ldots\ldots +\tan^{- 1}\left( \frac{2^{n- 1}}{1 +2^{2n- 1}} \right)+ \ldots\ldots is

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  4. Question 4

    26 Apr 2025 S2

    147
    If y=tan⁡−1(11+x+x2)+tan⁡−1(1x2+3x+3)+tan⁡−1(1x2+5x+7)y=\tan^{- 1}\left( \frac{1}{1 +x+x^{2}} \right)+\tan^{- 1}\left( \frac{1}{x^{2}+ 3x+ 3} \right)+\tan^{- 1}\left( \frac{1}{x^{2}+ 5x+ 7} \right) then the value of y′(0)y^{'}(0) is

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  5. Question 5

    18 Apr 2026 S1

    71
    If ∑n=12026tan⁡−1(1n2+n+1)=tan⁡−1(1−1x)\displaystyle\sum_{n=1}^{2026}\tan^{-1}\left(\dfrac{1}{n^2 + n + 1}\right) = \tan^{-1}\left(1 - \dfrac{1}{x}\right), where x≠0x \neq 0, then x=x =

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  6. Question 6

    20 Apr 2026 S1

    52
    If a1,a2,a3,…,ana_1, a_2, a_3, \ldots, a_n are in arithmetic progression with common difference d, then tan⁡[tan⁡−1(d1+a1a2)+tan⁡−1(d1+a2a3)+…+tan⁡−1(d1+an−1an)]=\tan\left[\tan^{-1}\left(\dfrac{d}{1 + a_1a_2}\right) + \tan^{-1}\left(\dfrac{d}{1 + a_2a_3}\right) + \ldots + \tan^{-1}\left(\dfrac{d}{1 + a_{n-1}a_n}\right)\right] = ____

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  7. Question 7

    14 May 2026 S2

    70
    The value of cot⁡−1(ab+1a−b)+tan⁡−1(b−cbc+1)+cot⁡−1(ac+1c−a)\cot^{-1}\left(\dfrac{ab + 1}{a - b}\right) + \tan^{-1}\left(\dfrac{b - c}{bc + 1}\right) + \cot^{-1}\left(\dfrac{ac + 1}{c - a}\right) is

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