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JEE Mains Maths · Inverse Trigonometric Functions

Simplifying Inverse Functions of a Variable

Simplifying an inverse function of an expression in x by a trigonometric substitution, turning compositions into algebra, and using the complementary identities.

All Inverse Trigonometric Functions notesJEE Mains Maths strategy

Why this matters

Fourteen PYQs, ten of them multiple choice, and one from 2026. Four simplify an inverse function of x by a trigonometric substitution; six turn a composition into algebra and then solve or eliminate; four combine sin⁻¹x + cos⁻¹x = π/2, or a sum of inverse sines equal to π, with a given condition. Three ideas cover the page.

Concept 1 of 3: Substitute and track the interval

An expression like 2x1−x22x\sqrt{1-x^2} is sin⁡2θ\sin2\theta once x=sin⁡θx=\sin\theta. Then sin⁡−1(sin⁡2θ)\sin^{-1}(\sin2\theta) equals 2θ2\theta only if 2θ2\theta lies in [−π2,π2]\left[-\frac\pi2,\frac\pi2\right]. The given interval for xx fixes the interval for θ\theta, and that decides the answer.

Definition

  • 1−x2\sqrt{1-x^2}: put x=sin⁡θx=\sin\theta or x=cos⁡θx=\cos\theta. 1+x2\sqrt{1+x^2}: put x=tan⁡θx=\tan\theta.
  • With x=tan⁡θx=\tan\theta: 2x1+x2=sin⁡2θ\frac{2x}{1+x^2}=\sin2\theta, 1−x21+x2=cos⁡2θ\frac{1-x^2}{1+x^2}=\cos2\theta, 2x1−x2=tan⁡2θ\frac{2x}{1-x^2}=\tan2\theta.
  • With x=sin⁡θx=\sin\theta: 3x−4x3=sin⁡3θ3x-4x^3=\sin3\theta; with x=cos⁡θx=\cos\theta: 2x2−1=cos⁡2θ2x^2-1=\cos2\theta.
  • Find θ\theta's interval from xx's, then bring the multiple of θ\theta into the principal range.

Standard substitutions

2tan⁡−1x=sin⁡−12x1+x2  (∣x∣≤1),2tan⁡−1x=cos⁡−11−x21+x2  (x≥0)2\tan^{-1}x=\sin^{-1}\frac{2x}{1+x^2}\ \ (|x|\le1),\qquad 2\tan^{-1}x=\cos^{-1}\frac{1-x^2}{1+x^2}\ \ (x\ge0)

Worked example

Simplify sin⁡−1(2x1−x2)\sin^{-1}\left(2x\sqrt{1-x^2}\right) for 12≤x≤1\frac1{\sqrt2}\le x\le1.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 4 April 2025 · Q63Moderate

Example 1 · Inverse Trigonometric Functions · Simplifying Inverse Functions of a Variable

Considering the principal values of the inverse trigonometric functions, sin⁡−1(32x+121−x2),−12<x<12\sin^{- 1}\left( \frac{\sqrt{3}}{2}x+\frac{1}{2}\sqrt{1 -x^{2}} \right), -\frac{1}{2}<x<\frac{1}{\sqrt{2}}, is equal to.

The interval for x decides the formula

sin⁡−1(2x1−x2)\sin^{-1}\left(2x\sqrt{1-x^2}\right) is 2sin⁡−1x2\sin^{-1}x only for ∣x∣≤12|x|\le\frac1{\sqrt2}. Outside that interval it is π−2sin⁡−1x\pi-2\sin^{-1}x or −π−2sin⁡−1x-\pi-2\sin^{-1}x. Read the interval before using a standard result.
Drill 3 more on substitute and track the interval

Concept 2 of 3: Turning a composition into algebra

A trigonometric function of an inverse function is an algebraic expression. For example, sin⁡(tan⁡−1x)=x1+x2\sin(\tan^{-1}x)=\frac{x}{\sqrt{1+x^2}}, read off a triangle with sides xx and 1. Replace each composition this way and the question becomes an equation or an identity in xx.

Definition

  • sin⁡(tan⁡−1x)=x1+x2\sin(\tan^{-1}x)=\frac{x}{\sqrt{1+x^2}}, cos⁡(tan⁡−1x)=11+x2\cos(\tan^{-1}x)=\frac1{\sqrt{1+x^2}}.
  • cos⁡(sin⁡−1x)=1−x2\cos(\sin^{-1}x)=\sqrt{1-x^2}, sin⁡(cos⁡−1x)=1−x2\sin(\cos^{-1}x)=\sqrt{1-x^2}.
  • sec⁡2(tan⁡−1x)=1+x2\sec^2(\tan^{-1}x)=1+x^2, csc⁡2(cot⁡−1x)=1+x2\csc^2(\cot^{-1}x)=1+x^2.
  • cos⁡(2sin⁡−1x)=1−2x2\cos(2\sin^{-1}x)=1-2x^2, sin⁡(2tan⁡−1x)=2x1+x2\sin(2\tan^{-1}x)=\frac{2x}{1+x^2}.

Composition to algebra

sin⁡(tan⁡−1x)=x1+x2,cos⁡(sin⁡−1x)=1−x2\sin(\tan^{-1}x)=\frac{x}{\sqrt{1+x^2}},\qquad \cos(\sin^{-1}x)=\sqrt{1-x^2}

Worked example

Solve cos⁡(tan⁡−1x)=sin⁡(cot⁡−134)\cos(\tan^{-1}x)=\sin\left(\cot^{-1}\frac34\right).
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 6 Apr 2026 Shift 2 · Q63Moderate

Example 2 · Inverse Trigonometric Functions · Simplifying Inverse Functions of a Variable

If sin⁡(tan⁡−1(x2))=cot⁡(sin⁡−11−x2),x∈(0,1)\sin\left( \tan^{- 1}(x\sqrt{2}) \right)= \cot\left( \sin^{- 1}\sqrt{1 -x^{2}} \right),x \in (0,1), then the value of x is :

Squaring brings extra roots

cos⁡(sin⁡−1x)=1−x2\cos(\sin^{-1}x)=\sqrt{1-x^2} is never negative, but tan⁡(cos⁡−1x)=1−x2x\tan(\cos^{-1}x)=\frac{\sqrt{1-x^2}}x takes the sign of xx. After squaring an equation, put each root back and check its sign.
Drill 5 more on turning a composition into algebra

Concept 3 of 3: Complementary identities with a condition

Since sin⁡−1x+cos⁡−1x=π2\sin^{-1}x+\cos^{-1}x=\frac\pi2, a second condition on the two angles pins each one down. A ratio condition splits π2\frac\pi2 in that ratio. A difference of squares factors as a sum times a difference, and the sum is π2\frac\pi2.

Definition

  • sin⁡−1x+cos⁡−1x=π2\sin^{-1}x+\cos^{-1}x=\frac\pi2 for ∣x∣≤1|x|\le1; the same for tan⁡−1\tan^{-1} and cot⁡−1\cot^{-1}.
  • Ratio: if sin⁡−1xa=cos⁡−1xb=k\frac{\sin^{-1}x}{a}=\frac{\cos^{-1}x}{b}=k, then k(a+b)=π2k(a+b)=\frac\pi2.
  • (sin⁡−1x)2−(cos⁡−1x)2=π2(sin⁡−1x−cos⁡−1x)(\sin^{-1}x)^2-(\cos^{-1}x)^2=\frac\pi2\left(\sin^{-1}x-\cos^{-1}x\right).
  • If sin⁡−1α+sin⁡−1β+sin⁡−1γ=π\sin^{-1}\alpha+\sin^{-1}\beta+\sin^{-1}\gamma=\pi, then α,β,γ\alpha,\beta,\gamma are the sines of the angles of a triangle, and the sine rule applies.

Complementary identity

sin⁡−1x+cos⁡−1x=π2,−1≤x≤1\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2},\quad -1\le x\le1

Worked example

If sin⁡−1x:cos⁡−1x=1:2\sin^{-1}x:\cos^{-1}x=1:2, find xx.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 31 January 2024 · Q75Moderate

Example 3 · Inverse Trigonometric Functions · Simplifying Inverse Functions of a Variable

For α,β,γ≠0\alpha,\beta,\gamma\neq 0. If sin⁡−1α+sin⁡−1β+sin⁡−1γ=π\sin^{- 1}\alpha+\sin^{- 1}\beta+\sin^{- 1}\gamma=\pi and (α+β+γ)(α−γ+β)=3αβ(\alpha+\beta+\gamma)(\alpha-\gamma+\beta) = 3\alpha\beta, then γ\gamma equal to

The identity needs the same argument

sin⁡−1x+cos⁡−1y=π2\sin^{-1}x+\cos^{-1}y=\frac\pi2 holds only when x=yx=y. And sec⁡−1x+csc⁡−1x=π2\sec^{-1}x+\csc^{-1}x=\frac\pi2 needs ∣x∣≥1|x|\ge1. Check the arguments match before replacing a sum by π2\frac\pi2.
Drill 3 more on complementary identities with a condition

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)

  • Substitute and track the interval

    Standard substitutions

    2tan⁡−1x=sin⁡−12x1+x2  (∣x∣≤1),2tan⁡−1x=cos⁡−11−x21+x2  (x≥0)2\tan^{-1}x=\sin^{-1}\frac{2x}{1+x^2}\ \ (|x|\le1),\qquad 2\tan^{-1}x=\cos^{-1}\frac{1-x^2}{1+x^2}\ \ (x\ge0)
  • Turning a composition into algebra

    Composition to algebra

    sin⁡(tan⁡−1x)=x1+x2,cos⁡(sin⁡−1x)=1−x2\sin(\tan^{-1}x)=\frac{x}{\sqrt{1+x^2}},\qquad \cos(\sin^{-1}x)=\sqrt{1-x^2}
  • Complementary identities with a condition

    Complementary identity

    sin⁡−1x+cos⁡−1x=π2,−1≤x≤1\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2},\quad -1\le x\le1

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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