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(tan^-1 x)^2 + (cot^-1 x)^2 = 5 pi^2/8: find x

The same question, asked 3 times: 13 May 2024 S2, 25 Apr 2025 S2, 16 Apr 2026 S2. Every wording below, oldest first.

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  1. Question 1

    13 May 2024 S2

    111
    If (tan⁡−1x)2+(cot⁡−1x)2=5π28\left(\tan^{-1}x\right)^{2}+\left(\cot^{-1}x\right)^{2}=\frac{5\pi^{2}}{8}, then the value of xx is:

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  2. Question 2

    25 Apr 2025 S2

    120
    If (tan⁡−1x)2+(cot⁡−1x)2=5π28\left( \tan^{- 1}x \right)^{2}+\left( \cot^{- 1}x \right)^{2}=\frac{5\pi^{2}}{8}, then x2+1=x^{2}+ 1 =

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  3. Question 3

    16 Apr 2026 S2

    67
    If (tan⁡−1x)2+(cot⁡−1x)2=5π28(\tan^{-1}x)^2 + (\cot^{-1}x)^2 = \dfrac{5\pi^2}{8}, then the value of xx is equal to...

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