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MathsMediumClass 11

Equations — Solving inverse trig equations

Inverse Trigonometric Functions

3

JEE Qs

8%

Hard

75

min

Always verify solutions by substituting them back into the original equation to ensure they satisfy the domain and principal value range of all inverse trigonometric functions.

🧮 Key Formulas

sin(sin^-1 x) = x, for x in [-1, 1]
sin^-1(sin x) = x, for x in [-pi/2, pi/2]
cos(cos^-1 x) = x, for x in [-1, 1]
cos^-1(cos x) = x, for x in [0, pi]
tan(tan^-1 x) = x, for x in R
tan^-1(tan x) = x, for x in (-pi/2, pi/2)
sin^-1 x + cos^-1 x = pi/2, for x in [-1, 1]
tan^-1 x + cot^-1 x = pi/2, for x in R
sec^-1 x + cosec^-1 x = pi/2, for |x| >= 1
sin^-1(-x) = -sin^-1 x, for x in [-1, 1]
cos^-1(-x) = pi - cos^-1 x, for x in [-1, 1]
tan^-1(-x) = -tan^-1 x, for x in R
tan^-1 x + tan^-1 y = tan^-1((x+y)/(1-xy)), if xy < 1
tan^-1 x + tan^-1 y = pi + tan^-1((x+y)/(1-xy)), if xy > 1 and x,y > 0
tan^-1 x + tan^-1 y = -pi + tan^-1((x+y)/(1-xy)), if xy > 1 and x,y < 0
tan^-1 x - tan^-1 y = tan^-1((x-y)/(1+xy))
2 tan^-1 x = tan^-1(2x/(1-x^2)), if |x| < 1
2 tan^-1 x = sin^-1(2x/(1+x^2)), if |x| <= 1
2 tan^-1 x = cos^-1((1-x^2)/(1+x^2)), if x >= 0
sin^-1 x = cos^-1(sqrt(1-x^2)) = tan^-1(x/sqrt(1-x^2)), if x in [0, 1]
cos^-1 x = sin^-1(sqrt(1-x^2)) = tan^-1(sqrt(1-x^2)/x), if x in (0, 1]

✅ Key Points for JEE

  • 1Always identify and adhere to the domain and principal value range of each inverse trigonometric function involved in the equation; solutions outside these must be rejected.
  • 2When using sum/difference formulas for `tan^-1 x + tan^-1 y`, carefully apply the correct formula variant based on the product `xy` and the signs of `x` and `y` to account for `pi` adjustments.
  • 3Converting all inverse trigonometric functions in an equation to a single type (e.g., `tan^-1`) often simplifies the problem, especially before applying sum/difference identities.
  • 4Be cautious when squaring both sides of an equation to remove square roots, as this can introduce extraneous solutions; always verify all potential solutions in the original equation.
  • 5For complex expressions, consider substitution (e.g., `x = sin theta`, `x = tan theta`) to transform the inverse trigonometric equation into a simpler trigonometric equation, then solve for `theta` and convert back.

⚠️ Common Mistakes

  • ✕Failing to check if the obtained solutions lie within the domain of the original inverse trigonometric functions or correspond to their principal value ranges.
  • ✕Incorrectly applying the `tan^-1 x + tan^-1 y` formula without considering the necessary `pi` adjustments when `xy > 1`.
  • ✕Assuming `sin^-1(sin x) = x` (or similar identities for other functions) without checking if `x` lies within the principal value range of the inverse function.
  • ✕Misinterpreting `cos^-1(-x)` as `-cos^-1 x` instead of the correct `pi - cos^-1 x`.

📝 Practice Questions

See all

Q20.If α > β > γ > 0, then the expression cot−1 {β (α−β) } + cot−1 {γ (β−γ) } + cot−1 {α (γ−α) } equal to : (1) π (2) 0 (3) π 2 −(α + β + γ) (4) 3π L.

2025·MCQMedium

Q21.Let S = {x : cos−1 x = π + sin−1 x + sin−1(2x + 1)}. Then ∑x∈ S(2x −1)2 is equal to ______.

2025·NumericalMedium

Q11.Using the principal values of the inverse trigonometric functions, the sum of the maximum and the minimum is : values of 16 ((sec−1 x)2 + (cosec−1 x)2) (1) 24π2 (2) 22π2 (3) 31π2 (4) 18π2

2025·MCQMedium

Q20.If π 2 ≤x ≤3π4 , then cos−1 ( 1213 cos x + 135 sin x) is equal to (1) x −tan−1 43 (2) x + tan−1 45 (3) x −tan−1 125 (4) x + tan−1 125

2025·MCQMedium

Q10. cos (sin−1 35 + sin−1 135 + sin−1 3365 ) is equal to: (1) 1 (2) 0 (3) 32 (4) 33 65 65

2025·MCQMedium

Q86.Let the inverse trigonometric functions take principal values. The number of real solutions of the equation 2 sin−1 x + 3 cos−1 x = 2π5 , is _______

2024·NumericalEasy

NCERT Chapters

  • Class 12 Maths Ch 2: Inverse Trigonometric Functions