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Simplify inverse-trig of sqrt(1 + t) and sqrt(1 - t) ratios by a half-angle substitution

This type of question, asked 3 times: 11 May 2023 S1, 9 May 2024 S2, 16 Apr 2026 S2. All 3 questions below, oldest first.

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  1. Question 1

    11 May 2023 S1

    126
    The value of tan⁡−1(1+x−1−x1+x+1−x)+12cos⁡−1x\tan^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right) + \frac{1}{2}\cos^{-1}x is

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  2. Question 2

    9 May 2024 S2

    103
    The value of tan⁡−1 ⁣ ⁣(1+x2+1−x21+x2−1−x2),  ∣x∣<12,  x≠0\tan^{-1}\!\!\left(\dfrac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right),\;|x| < \frac{1}{\sqrt{2}},\; x \neq 0

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  3. Question 3

    16 Apr 2026 S2

    66
    The value of cot⁡−1[1−sin⁡x+1+sin⁡x1−sin⁡x−1+sin⁡x]\cot^{-1}\left[\dfrac{\sqrt{1 - \sin x} + \sqrt{1 + \sin x}}{\sqrt{1 - \sin x} - \sqrt{1 + \sin x}}\right], where x∈(0,π2)x \in \left(0, \dfrac{\pi}{2}\right) is...

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