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Solve an equation with a sum of two tan^-1 terms using the addition formula

This type of question, asked 8 times: 2 May 2023 S1, 4 May 2023 S1, 10 May 2023 S2, 16 May 2023 S2, 10 May 2024 S1, 22 Apr 2025 S2, 25 Apr 2025 S2, 18 Apr 2026 S2. All 9 questions below, oldest first.

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  1. Question 1

    2 May 2023 S1

    106
    Considering only the principal values of inverse function, the set A={x≥0 | tan⁡−1(2x)+tan⁡−1(3x)=π4}A = \left\{x \ge 0\,\middle|\, \tan^{-1}(2x) + \tan^{-1}(3x) = \frac{\pi}{4}\right\}

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  2. Question 2

    4 May 2023 S1

    124
    Considering only the Principal values of inverse functions, the set A={x≥0∣tan⁡−1(2x)+tan⁡−1(3x)=π4}A = \{x\geq0 \mid \tan^{-1}(2x)+\tan^{-1}(3x)=\frac{\pi}{4}\}

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  3. Question 3

    10 May 2023 S2

    143
    If tan⁡−1(x+2)+tan⁡−1(x−2)−tan⁡−112=0\tan^{-1}(x+2)+\tan^{-1}(x-2)-\tan^{-1}\frac{1}{2}=0, then one value of x is

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  4. Question 4

    16 May 2023 S2

    136
    If tan⁡−1x+1x−1+tan⁡−1x−1x=tan⁡−1(−7)\tan^{-1}\frac{x+1}{x-1} + \tan^{-1}\frac{x-1}{x} = \tan^{-1}(-7), then xx is equal to

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  5. Question 5

    10 May 2024 S1

    103
    Solution of tan⁡−1(1+x)+tan⁡−1(1−x)=π2\tan^{-1}(1+x)+\tan^{-1}(1-x)=\frac{\pi}{2} is

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  6. Question 6

    10 May 2024 S1

    137
    Set A={x≥0∣tan⁡−1x+tan⁡−1(6x)=π4}A=\{x\ge0\mid\tan^{-1}x+\tan^{-1}(6x)=\frac{\pi}{4}\}

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  7. Question 7

    22 Apr 2025 S2

    132
    The number of solutions of tan⁡−1(x+2x)−tan⁡−1(4x)−tan⁡−1(x−2x)=0\tan^{- 1}\left( x+\frac{2}{x} \right)-\tan^{- 1}\left( \frac{4}{x} \right)-\tan^{- 1}\left( x-\frac{2}{x} \right)= 0 are

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  8. Question 8

    25 Apr 2025 S2

    135
    If tan⁡−1(x+1)+tan⁡−1x+tan⁡−1(x−1)=tan⁡−13\tan^{- 1}(x+ 1) +\tan^{- 1}x+\tan^{- 1}(x- 1) =\tan^{- 1}3, then for x<0x< 0 the value of 500x4+270x2+997=500x^{4}+ 270x^{2}+ 997 =

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  9. Question 9

    18 Apr 2026 S2

    67
    If tan⁡−1ax+tan⁡−13x=π4\tan^{-1}ax + \tan^{-1}3x = \dfrac{\pi}{4}, where 3ax2<13ax^2 < 1, then value of aa for x=16x = \dfrac{1}{6} is …\ldots

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