Last Updated: May 2026
Inverse Trigonometric Functions appear in roughly 1–2 questions in every JEE Main 2027 paper and frequently in JEE Advanced as combined problems with calculus. The chapter (NCERT Class 12 Chapter 2) tests domain-range mastery, identity manipulation, and the principal-value branch convention.
Quick Reference Table — Domain, Range, Principal Branch
| Function | Domain | Principal Range |
|---|---|---|
| sin⁻¹ x | [−1, 1] | [−π/2, π/2] |
| cos⁻¹ x | [−1, 1] | [0, π] |
| tan⁻¹ x | (−∞, ∞) | (−π/2, π/2) |
| cot⁻¹ x | (−∞, ∞) | (0, π) |
| sec⁻¹ x | (−∞, −1] ∪ [1, ∞) | [0, π] − {π/2} |
| cosec⁻¹ x | (−∞, −1] ∪ [1, ∞) | [−π/2, π/2] − {0} |
Six Foundational Identities
- sin⁻¹(−x) = −sin⁻¹ x ; tan⁻¹(−x) = −tan⁻¹ x ; cosec⁻¹(−x) = −cosec⁻¹ x (odd functions)
- cos⁻¹(−x) = π − cos⁻¹ x ; cot⁻¹(−x) = π − cot⁻¹ x ; sec⁻¹(−x) = π − sec⁻¹ x
- sin⁻¹ x + cos⁻¹ x = π/2 ; tan⁻¹ x + cot⁻¹ x = π/2 ; sec⁻¹ x + cosec⁻¹ x = π/2 for all x in respective domain
- tan⁻¹ x + tan⁻¹ y = tan⁻¹[(x + y)/(1 − xy)] when xy < 1
- tan⁻¹ x − tan⁻¹ y = tan⁻¹[(x − y)/(1 + xy)]
- 2 tan⁻¹ x = tan⁻¹[2x/(1 − x²)] = sin⁻¹[2x/(1 + x²)] = cos⁻¹[(1 − x²)/(1 + x²)]
Domain-Restriction Trap
The identity tan⁻¹ x + tan⁻¹ y = tan⁻¹[(x + y)/(1 − xy)] is only true when xy < 1. When xy > 1 and x, y > 0, add π. When xy > 1 and x, y < 0, subtract π. JEE Main loves to plant traps where the candidate forgets this caveat.
Three High-Yield Question Types
- Find the principal value — express the inverse trig in [−π/2, π/2] etc. given an angle outside the principal range
- Simplify a long expression — combine multiple inverse trig terms into a single function using identities 4, 5, 6
- Differentiation / integration involving inverse trig — d(sin⁻¹ x)/dx = 1/√(1−x²), d(tan⁻¹ x)/dx = 1/(1+x²), etc.
Common Compositions
- sin(sin⁻¹ x) = x for x ∈ [−1, 1]
- sin⁻¹(sin x) = x only when x ∈ [−π/2, π/2]; else reduce
- cos(cos⁻¹ x) = x for x ∈ [−1, 1]
- cos⁻¹(cos x) = x only when x ∈ [0, π]; else reduce
Reduction trick: for sin⁻¹(sin x) where x is outside [−π/2, π/2], find the equivalent angle in this range using sin x = sin(π − x) = sin(2π + x).
Useful Algebraic Substitutions
- If x = sin θ, √(1 − x²) = cos θ → simplifies sin⁻¹ √(1−x²) integrals
- If x = tan θ, 1 + x² = sec² θ → simplifies tan⁻¹ √(1 + x²) terms
- If x = sec θ, x² − 1 = tan² θ → useful for sec⁻¹ √(x² − 1)
35 Practice MCQs — JEE Main Inverse Trigonometric Functions
[cg_quiz id=”cg-jee-inv-trig-2027″]
Frequently Asked Questions
What is the principal value of sin⁻¹(sin 5π/6)?
5π/6 is outside [−π/2, π/2]. Use sin(5π/6) = sin(π − 5π/6) = sin(π/6). So sin⁻¹(sin 5π/6) = π/6.
When is the formula tan⁻¹ x + tan⁻¹ y = tan⁻¹[(x+y)/(1−xy)] invalid?
When xy ≥ 1. If both x, y > 0 and xy > 1, add π to the RHS. If both x, y < 0 and xy > 1, subtract π.
What is sin⁻¹ x + cos⁻¹ x?
π/2 for all x ∈ [−1, 1]. The two functions are complementary.
What is the derivative of tan⁻¹ x?
d(tan⁻¹ x)/dx = 1 / (1 + x²) for all real x.
Continue Your JEE 2027 Prep
- JEE Main Coordinate Geometry 2027
- JEE Main Complex Numbers 2027
- JEE Gurukul Courses
- JEE Free Mock Test
Bottom line: Memorise the principal-range table and the six identities. Watch the xy < 1 trap on tan⁻¹ addition. Practise sin⁻¹(sin x) reduction for arguments outside [−π/2, π/2].