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JEE Mains Maths · Inverse Trigonometric Functions

Domain, Range and Principal Values

Where each inverse function is defined, what values it can take, and how an inverse of a trigonometric value is brought back into the principal range.

All Inverse Trigonometric Functions notesJEE Mains Maths strategy

Why this matters

Sixteen PYQs, fourteen of them multiple choice, and one from 2026. Six find the domain of an inverse function of an expression; four find a range, a maximum or a minimum; six evaluate inverse functions of trigonometric values, where the principal range decides the answer. Three ideas cover the page.

Concept 1 of 3: Domain conditions on the argument

Each inverse function accepts only certain arguments. sin⁡−1\sin^{-1} and cos⁡−1\cos^{-1} need the argument in [−1,1][-1,1]; sec⁡−1\sec^{-1} and csc⁡−1\csc^{-1} need it outside (−1,1)(-1,1); tan⁡−1\tan^{-1} and cot⁡−1\cot^{-1} take anything. So the domain of sin⁡−1(g(x))\sin^{-1}(g(x)) is the set of xx with −1≤g(x)≤1-1\le g(x)\le1: an inequality in xx.

Definition

  • sin⁡−1u, cos⁡−1u\sin^{-1}u,\ \cos^{-1}u: −1≤u≤1-1\le u\le1.
  • sec⁡−1u, csc⁡−1u\sec^{-1}u,\ \csc^{-1}u: ∣u∣≥1|u|\ge1.
  • tan⁡−1u, cot⁡−1u\tan^{-1}u,\ \cot^{-1}u: every real uu.
  • For ∣pq∣≤1\left|\frac{p}{q}\right|\le1, solve p2≤q2p^2\le q^2 with q≠0q\ne0.
  • For a sum of terms, intersect their domains.

Domains

sin⁡−1u, cos⁡−1u: ∣u∣≤1,sec⁡−1u, csc⁡−1u: ∣u∣≥1\sin^{-1}u,\ \cos^{-1}u:\ |u|\le1,\qquad \sec^{-1}u,\ \csc^{-1}u:\ |u|\ge1

Worked example

Find the domain of cos⁡−1(x+1x−1)\cos^{-1}\left(\frac{x+1}{x-1}\right).
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 10 April 2023 · Q177Moderate

Example 1 · Inverse Trigonometric Functions · Domain, Range and Principal Values

If the domain of the function f(x)=sec⁡−1(2x5x+3)f(x) =\sec^{- 1}\left( \frac{2x}{5x + 3} \right) is [α,β)∪(γ,δ]\lbrack\alpha,\beta) \cup (\gamma,\delta\rbrack, then ∣3α+10(β+γ)+21δ∣|3\alpha + 10(\beta + \gamma) + 21\delta| is equal to

Do not multiply by a denominator of unknown sign

Turning pq≤1\frac{p}{q}\le1 into p≤qp\le q is wrong when q<0q<0. Square instead: ∣pq∣≤1\left|\frac pq\right|\le1 exactly when p2≤q2p^2\le q^2, and then remove the points where q=0q=0.
Drill 5 more on domain conditions on the argument

Concept 2 of 3: Ranges and complementary pairs

To find a range, first find the set of values the argument takes, then push it through the inverse function, which is monotonic. When an expression mixes sin⁡−1x\sin^{-1}x and cos⁡−1x\cos^{-1}x, replace one with π2\frac\pi2 minus the other. The expression becomes a polynomial in one angle aa, and aa runs over a known interval.

Definition

  • Ranges: sin⁡−1\sin^{-1}: [−π2,π2]\left[-\frac\pi2,\frac\pi2\right]; cos⁡−1\cos^{-1}: [0,π][0,\pi]; tan⁡−1\tan^{-1}: (−π2,π2)\left(-\frac\pi2,\frac\pi2\right); cot⁡−1\cot^{-1}: (0,π)(0,\pi).
  • sec⁡−1\sec^{-1}: [0,π]−{π2}[0,\pi]-\left\{\frac\pi2\right\}; csc⁡−1\csc^{-1}: [−π2,π2]−{0}\left[-\frac\pi2,\frac\pi2\right]-\{0\}.
  • sin⁡−1x+cos⁡−1x=π2\sin^{-1}x+\cos^{-1}x=\frac\pi2 for ∣x∣≤1|x|\le1; tan⁡−1x+cot⁡−1x=π2\tan^{-1}x+\cot^{-1}x=\frac\pi2 for all xx; sec⁡−1x+csc⁡−1x=π2\sec^{-1}x+\csc^{-1}x=\frac\pi2 for ∣x∣≥1|x|\ge1.
  • With a=sin⁡−1xa=\sin^{-1}x, a2+(π2−a)2a^2+\left(\frac\pi2-a\right)^2 is a parabola in aa: least at a=π4a=\frac\pi4, greatest at the end of the interval farther from π4\frac\pi4.

Complementary pairs

sin⁡−1x+cos⁡−1x=tan⁡−1x+cot⁡−1x=π2\sin^{-1}x+\cos^{-1}x=\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}

Worked example

Find the range of (tan⁡−1x)2+(cot⁡−1x)2(\tan^{-1}x)^2+(\cot^{-1}x)^2.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 21 Jan 2026 Shift 2 · Q73Moderate

Example 2 · Inverse Trigonometric Functions · Domain, Range and Principal Values

Let the maximum value of (sin⁡−1x)2+(cos⁡−1x)2\left( \sin^{- 1}x \right)^{2} + \left( \cos^{- 1}x \right)^{2} for x∈[−32,12]x \in \left\lbrack - \frac{\sqrt{3}}{2},\frac{1}{\sqrt{2}} \right\rbrack be mnπ2\frac{m}{n}\pi^{2}, where gcd(m,n)=1gcd(m,n) = 1. Then m+nm + n is equal to ____\_\_\_\_ .

Check whether each endpoint is reached

tan⁡−1x\tan^{-1}x never equals ±π2\pm\frac\pi2, and x2x2+1\frac{x^2}{x^2+1} never equals 1. An endpoint that is only approached gets a round bracket, and options often differ only in that bracket.
Drill 3 more on ranges and complementary pairs

Concept 3 of 3: Inverse of a trigonometric value outside the principal range

sin⁡−1(sin⁡x)\sin^{-1}(\sin x) is the angle in [−π2,π2]\left[-\frac\pi2,\frac\pi2\right] with the same sine as xx. It equals xx only when xx is already in that interval. Otherwise, use sin⁡(π−x)=sin⁡x\sin(\pi-x)=\sin x or the period 2π2\pi to move xx into the range. Do the same for cos⁡−1(cos⁡x)\cos^{-1}(\cos x) on [0,π][0,\pi] and tan⁡−1(tan⁡x)\tan^{-1}(\tan x) on (−π2,π2)\left(-\frac\pi2,\frac\pi2\right).

Definition

  • sin⁡−1(sin⁡x)=x\sin^{-1}(\sin x)=x on [−π2,π2]\left[-\frac\pi2,\frac\pi2\right]; =π−x=\pi-x on [π2,3π2]\left[\frac\pi2,\frac{3\pi}2\right]; =x−2π=x-2\pi on [3π2,5π2]\left[\frac{3\pi}2,\frac{5\pi}2\right].
  • cos⁡−1(cos⁡x)=x\cos^{-1}(\cos x)=x on [0,π][0,\pi]; =2π−x=2\pi-x on [π,2π][\pi,2\pi]; cos⁡−1(cos⁡(−x))=cos⁡−1(cos⁡x)\cos^{-1}(\cos(-x))=\cos^{-1}(\cos x).
  • tan⁡−1(tan⁡x)=x−kπ\tan^{-1}(\tan x)=x-k\pi, with kk chosen so the result lies in (−π2,π2)\left(-\frac\pi2,\frac\pi2\right).
  • Use π≈3.14\pi\approx3.14: π2≈1.57\frac\pi2\approx1.57, 3π2≈4.71\frac{3\pi}2\approx4.71, 2π≈6.282\pi\approx6.28.

Back into the principal range

sin⁡−1(sin⁡x)=π−x  (π2≤x≤3π2),cos⁡−1(cos⁡x)=2π−x  (π≤x≤2π)\sin^{-1}(\sin x)=\pi-x\ \ \left(\tfrac{\pi}{2}\le x\le\tfrac{3\pi}{2}\right),\qquad \cos^{-1}(\cos x)=2\pi-x\ \ (\pi\le x\le2\pi)

Worked example

Find sin⁡−1(sin⁡4)+cos⁡−1(cos⁡4)\sin^{-1}(\sin4)+\cos^{-1}(\cos4).
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 31 January 2024 · Q166Moderate

Example 3 · Inverse Trigonometric Functions · Domain, Range and Principal Values

If a=sin⁡−1(sin⁡(5))a =\sin^{- 1}(\sin(5)) and b=cos⁡−1(cos⁡(5))b =\cos^{- 1}(\cos(5)), then a2+b2a^{2}+b^{2} is equal to

sin⁻¹(sin x) is not always x

sin⁡−1(sin⁡3)\sin^{-1}(\sin3) is π−3\pi-3, not 3, because 3 lies outside [−π2,π2]\left[-\frac\pi2,\frac\pi2\right]. Before writing the answer, check that it lies in the principal range.
Drill 5 more on inverse of a trigonometric value outside the principal range

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

  • Domain conditions on the argument

    Domains

    sin⁡−1u, cos⁡−1u: ∣u∣≤1,sec⁡−1u, csc⁡−1u: ∣u∣≥1\sin^{-1}u,\ \cos^{-1}u:\ |u|\le1,\qquad \sec^{-1}u,\ \csc^{-1}u:\ |u|\ge1
  • Ranges and complementary pairs

    Complementary pairs

    sin⁡−1x+cos⁡−1x=tan⁡−1x+cot⁡−1x=π2\sin^{-1}x+\cos^{-1}x=\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}
  • Inverse of a trigonometric value outside the principal range

    Back into the principal range

    sin⁡−1(sin⁡x)=π−x  (π2≤x≤3π2),cos⁡−1(cos⁡x)=2π−x  (π≤x≤2π)\sin^{-1}(\sin x)=\pi-x\ \ \left(\tfrac{\pi}{2}\le x\le\tfrac{3\pi}{2}\right),\qquad \cos^{-1}(\cos x)=2\pi-x\ \ (\pi\le x\le2\pi)

Watch out for (3)

Test yourself on Inverse Trigonometric Functions

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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