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Simplify expressions in 2x/(1+x^2) and (1-x^2)/(1+x^2) with x = tan t

This type of question, asked 4 times: 21 Apr 2025 S2, 22 Apr 2025 S1, 20 Apr 2026 S2, 15 May 2026 S2. All 4 questions below, oldest first.

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  1. Question 1

    21 Apr 2025 S2

    105
    The value of sin⁡[tan⁡−1(1−x22x)+cos⁡−1(1−x21+x2)]\sin\left\lbrack \tan^{- 1}\left( \frac{1 -x^{2}}{2x} \right)+\cos^{- 1}\left( \frac{1 -x^{2}}{1 +x^{2}} \right) \right\rbrack is

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  2. Question 2

    22 Apr 2025 S1

    146
    If 3sin⁡−1(2x1+x2)−4cos⁡−1(1−x21+x2)+2tan⁡−1(2x1−x2)=π33\sin^{- 1}\left( \frac{2x}{1 +x^{2}} \right)- 4\cos^{- 1}\left( \frac{1 -x^{2}}{1 +x^{2}} \right)+ 2\tan^{- 1}\left( \frac{2x}{1 -x^{2}} \right)=\frac{\pi}{3} then the value of x=x=

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  3. Question 3

    20 Apr 2026 S2

    66
    If 3sin⁡−1(2x1+x2)−4cos⁡−1(1−x21+x2)+2tan⁡−1(2x1−x2)=π33\sin^{-1}\left(\dfrac{2x}{1 + x^2}\right) - 4\cos^{-1}\left(\dfrac{1 - x^2}{1 + x^2}\right) + 2\tan^{-1}\left(\dfrac{2x}{1 - x^2}\right) = \dfrac{\pi}{3} then x=x =?

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  4. Question 4

    15 May 2026 S2

    70
    If x∈[0,1)x \in [0, 1) then the value of tan⁡{12sin⁡−1(2x1+x2)+12cos⁡−1(1−x21+x2)}\tan\left\{\dfrac{1}{2}\sin^{-1}\left(\dfrac{2x}{1 + x^2}\right) + \dfrac{1}{2}\cos^{-1}\left(\dfrac{1 - x^2}{1 + x^2}\right)\right\} is:

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