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JEE Main Inverse Trigonometric Functions questions

12 Mathematics questions on Inverse Trigonometric Functions as asked in JEE Main, newest first. 9 come with a worked solution.

  1. tan^-1 (dfrac1+√(3)3+√(3) ) + sec^-1 (sqrtdfrac8+4√(3)6+3√(3) ) is equal to :

    2023 · with solution

  2. All x satisfying the inequality (cot^-1 x)^2 - 7(cot^-1 x) + 10 > 0, lie in the interval

    2019

  3. The value of sin^-1 (12/13 ) - sin^-1 (3/5 ) is equal to

    2019

  4. If cos^-1x - cos^-1y/2 = α, where -1 ≤ x ≤ 1, -2 ≤ y ≤ 2, x ≤ y/2, then for all x, y, 4x^2 - 4xycosα + y^2 is equal to

    2019 · with solution

  5. If α = cos^-1 (3/5 ), β = tan^-1 (1/3 ), where 0 < α, β < (π)/2, then α - β is equal to

    2019 · with solution

  6. Considering only the principal values of inverse functions, the set A = \x ≥ 0: tan^-1(2x) + tan^-1(3x) = (π)/4 \

    2019

  7. If x = sin^-1(sin 10) and y = cos^-1(cos 10), then y - x is equal to

    2019 · with solution

  8. The value of cot \sum_n=1^23 cot^-1 (1 + sum_k=1^n 2k ) \ is

    2013 · with solution

  9. If x, y and z are in AP and tan^-1 x, tan^-1 y and tan^-1 z are also in AP, then

    2013 · with solution

  10. The number of real solutions of tan^-1√(x(x+1)) + sin^-1√(x^2+x+1) = (π)/2 is

    1999 · with solution

  11. The value of tan [cos^-1 (4/5 ) + tan^-1 (2/3 ) ] is

    1983 · with solution

  12. If tan^-1 y = tan^-1 x + tan^-1 (2x/(1-x^2) ), where |x| < dfrac1√(3). Then, the value of y is

    with solution

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