Inverse Trigonometry for JEE Main: Identities & Equations

Inverse trigonometry contributes two to three marks to JEE Main Mathematics every session and is one of the chapters where errors come almost exclusively from domain/range confusion rather than algebraic mistakes. The student who internalises the principal value ranges for all six inverse functions and knows when an identity applies within that range will answer every inverse trig question correctly. This guide builds that foundation and then covers all the identities JEE actually tests.

Principal Value Ranges and Domain

Each inverse trigonometric function has a restricted range to ensure it is a true function: sin⁻¹: domain [−1,1], range [−π/2, π/2]; cos⁻¹: domain [−1,1], range [0, π]; tan⁻¹: domain ℝ, range (−π/2, π/2); cosec⁻¹: domain (−∞,−1]∪[1,∞), range [−π/2, π/2]{0}; sec⁻¹: domain (−∞,−1]∪[1,∞), range [0, π]{π/2}; cot⁻¹: domain ℝ, range (0, π).

JEE uses these ranges in three ways: (1) asking which value sin⁻¹(sin 5π/6) equals — since 5π/6 is outside [−π/2, π/2], you must find the equivalent value in the principal range: sin 5π/6 = sin π/6 = 1/2, so sin⁻¹(sin 5π/6) = π/6; (2) domain questions asking where a composite inverse expression is defined; (3) simplification questions where you must identify whether a given inverse is in the principal range. Take a free inverse trig mock to practise principal value identification under time pressure.

Key Identities and When They Apply

sin⁻¹(−x) = −sin⁻¹x; cos⁻¹(−x) = π − cos⁻¹x; tan⁻¹(−x) = −tan⁻¹x. Note the asymmetry: sin⁻¹ and tan⁻¹ are odd functions; cos⁻¹ is not. sin⁻¹x + cos⁻¹x = π/2 for x ∈ [−1,1]; tan⁻¹x + cot⁻¹x = π/2 for x ∈ ℝ; sec⁻¹x + cosec⁻¹x = π/2 for |x| ≥ 1.

The addition formula for tan⁻¹: tan⁻¹x + tan⁻¹y = tan⁻¹((x+y)/(1−xy)) if xy < 1; = π + tan⁻¹((x+y)/(1−xy)) if xy > 1 and x > 0; = −π + tan⁻¹((x+y)/(1−xy)) if xy > 1 and x < 0. The xy < 1 condition is the critical gate: applying the simple formula when xy > 1 gives the wrong answer (missing the π adjustment). JEE exploits this condition boundary repeatedly. Subtraction: tan⁻¹x − tan⁻¹y = tan⁻¹((x−y)/(1+xy)), with analogous conditions.

Composition of Trigonometric and Inverse Functions

sin(sin⁻¹x) = x for x ∈ [−1,1]. sin⁻¹(sinx) = x only if x ∈ [−π/2, π/2]; otherwise it must be reflected to the principal range. These compositions appear in JEE as simplification questions. For compound expressions like sin(2 sin⁻¹ x): let θ = sin⁻¹x, so sinθ = x, cosθ = √(1−x²), and sin(2θ) = 2x√(1−x²). This "let θ = inverse" substitution converts any compound expression into a straightforward trigonometric calculation.

Multiple angle: 2 tan⁻¹x = sin⁻¹(2x/(1+x²)) for |x| ≤ 1; = cos⁻¹((1−x²)/(1+x²)) for x ≥ 0. These relations connect the inverse tangent to the other inverse functions and are tested in "simplify" or "prove" questions. The formulae for 3 tan⁻¹x and tan⁻¹(1/x) = π/2 − tan⁻¹x (for x > 0) complete the set needed for JEE. Pair this guide with our trigonometric functions guide and limits and continuity guide for the full analysis of trig-calculus composite questions.