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Inverse Trigonometric Functions — Class 12 solved problems

10 problems from this chapter, each solved step by step.

  1. cos ( sin^(-1) (3)/(8) + sin^(-1) (5)/(13) + sin^(-1) (33)/(65)) is equal to:

    (3√55 - 12)/(40)

  2. Considering the principal values of the inverse trigonometric functions, sin^(-1) ((√3)/(2)x + (1)/(2)√(1-x²)), -(1)/(2) < x < (1)/(√2) is equal to:

    (π)/(6) + sin^(-1)x

  3. If α > β > > 0, then the expression cot^(-1) β + ((1 + β^5))/((α - β)) + cot^(-1) + ((1 + ²))/((β -)) + cot^(-1) α + ((1 + α²))/(( - α)) is equal to:

    0

  4. The value of cot^(-1) ( (√(1 + tan²(2)) - 1)/(tan(2))) - cot^(-1) ( (√(1 + tan²((1)/(2))) + 1)/(tan((1)/(2)))) is equal to:

    (π)/(2) - (5)/(4)

  5. Solve for x: sin⁻¹(x) + sin⁻¹(√(1-x²)) = π/2 for x ∈ [0,1].

    The solution for x is x [0,1].

  6. Using the principal values of the inverse trigonometric functions, the sum of the maximum and the minimum values of 16((sec^(-1) x)² + (csc^(-1) x)²) is:

    22π²

  7. If for some α, β; α ≤ β, α + β = 8 and sec²(tan^(-1) α) + csc²(cot^(-1) β) = 36, then α² + β is

    14

  8. If y = cos((π)/(3) + cos^(-1)(x)/(2)), then (x - y)² + 3y² is equal to:

    3

  9. If (π)/(2) ≤ x ≤ (3π)/(4), then cos^(-1) ( (12)/(13) cos x + (5)/(13) sin x) is equal to

    x - tan^(-1)((5)/(12))

  10. Find the value of: tan^(-1)(2 * cos(2 * sin^(-1)(1/2)))

    The value of tan^(-1)(2 · cos(2 · sin^(-1)(1/2))) is (π)/(4).

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