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JEE Mains Maths · Inverse Trigonometric Functions

Equations in Inverse Trigonometric Functions

Solving equations in inverse functions by combining terms and taking a tangent, sine or cosine, then checking each root against the principal ranges.

All Inverse Trigonometric Functions notesJEE Mains Maths strategy

Why this matters

Sixteen PYQs, ten of them multiple choice, and two from 2026. Six are equations in tan⁻¹ and cot⁻¹, or become one; seven are equations in sin⁻¹ and cos⁻¹, solved by taking a sine or cosine; three are settled by the domain alone. Three ideas cover the page.

Concept 1 of 3: Equations in inverse tangents

Combine the inverse tangents with the addition formula, take the tangent of both sides, and solve the resulting polynomial. Taking the tangent forgets which quadrant the angle was in, so some roots of the polynomial do not satisfy the original equation. Put every root back.

Definition

  • tan⁡−1a+tan⁡−1b=c\tan^{-1}a+\tan^{-1}b=c: take tangents, a+b1−ab=tan⁡c\frac{a+b}{1-ab}=\tan c.
  • cot⁡−1x=tan⁡−11x\cot^{-1}x=\tan^{-1}\frac1x only for x>0x>0; for x<0x<0 it is π+tan⁡−11x\pi+\tan^{-1}\frac1x.
  • A root is valid only if the left side, with principal values, really equals the right side.
  • For a count of solutions, also check the stated interval for xx.

Taking tangents

tan⁡−1a+tan⁡−1b=c ⇒ a+b1−ab=tan⁡c\tan^{-1}a+\tan^{-1}b=c\ \Rightarrow\ \frac{a+b}{1-ab}=\tan c

Worked example

Solve tan⁡−12x+tan⁡−13x=π4\tan^{-1}2x+\tan^{-1}3x=\frac\pi4.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 22 Jan 2026 Shift 1 · Q68Moderate

Example 1 · Inverse Trigonometric Functions · Equations in Inverse Trigonometric Functions

The number of solutions of tan⁡−14x+tan⁡−16x=π6\tan^{- 1}4x +\tan^{- 1}6x =\frac{\pi}{6}, where −126<x<126-\frac{1}{2\sqrt{6}}< x <\frac{1}{2\sqrt{6}} is equal to

Taking tangents loses the range

tan⁡(A+B)=1\tan(A+B)=1 also holds when A+B=−3π4A+B=-\frac{3\pi}4. Every root of the polynomial must be put back into the original equation; a negative root often fails.
Drill 5 more on equations in inverse tangents

Concept 2 of 3: Equations in inverse sines and cosines

First use cos⁡−1x=π2−sin⁡−1x\cos^{-1}x=\frac\pi2-\sin^{-1}x to reduce the number of different functions. Isolate one inverse function, then take the sine or cosine of both sides. The result is algebraic, often after squaring. Squaring and taking sines both create extra roots, so check each root against the ranges.

Definition

  • Replace cos⁡−1x\cos^{-1}x by π2−sin⁡−1x\frac\pi2-\sin^{-1}x to leave one function where possible.
  • If sin⁡−1u=θ\sin^{-1}u=\theta, then u=sin⁡θu=\sin\theta and θ\theta must lie in [−π2,π2]\left[-\frac\pi2,\frac\pi2\right].
  • A side that must lie in a range gives a sign condition on xx; use it to reject roots.
  • If the reduced equation asks for cos⁡−1x\cos^{-1}x outside [0,π][0,\pi], there is no solution.

Isolate, then take the sine

sin⁡−1u=θ ⇒ u=sin⁡θ,−π2≤θ≤π2\sin^{-1}u=\theta\ \Rightarrow\ u=\sin\theta,\quad -\tfrac{\pi}{2}\le\theta\le\tfrac{\pi}{2}

Worked example

Solve sin⁡−1(1−x)−2sin⁡−1x=π2\sin^{-1}(1-x)-2\sin^{-1}x=\frac\pi2.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 29 Jan 2025 · Q75Moderate

Example 2 · Inverse Trigonometric Functions · Equations in Inverse Trigonometric Functions

Let S={x:cos⁡−1x=π+sin⁡−1x+sin⁡−1(2x+1)}S=\{x:\cos^{-1}x=\pi+\sin^{-1}x+\sin^{-1}(2x+1)\}. Then ∑x∈S(2x−1)2\sum_{x\in S}(2x-1)^{2} is equal to

Squaring adds roots

Taking a sine and squaring keeps every true root but can add false ones. In the worked example x=12x=\frac12 solves the quadratic but not the equation. Check each root in the original equation, with principal values.
Drill 6 more on equations in inverse sines and cosines

Concept 3 of 3: When the domain pins x down

Sometimes the conditions for the terms to exist leave only a few values of xx. sin⁡−1u\sin^{-1}\sqrt{u} needs 0≤u≤10\le u\le1; if another term needs u≥1u\ge1, then u=1u=1 exactly. Find those few values first, then test each in the equation. No trigonometric manipulation is needed.

Definition

  • sin⁡−1u, cos⁡−1u\sin^{-1}u,\ \cos^{-1}u: ∣u∣≤1|u|\le1. u\sqrt u: u≥0u\ge0. sec⁡−1u\sec^{-1}u: ∣u∣≥1|u|\ge1.
  • Two conditions such as u≥1u\ge1 and u≤1u\le1 force u=1u=1: a few values of xx.
  • sin⁡−1[t]\sin^{-1}[t] with the greatest integer: [t]∈{−1,0,1}[t]\in\{-1,0,1\}, so the value is −π2-\frac\pi2, 0 or π2\frac\pi2.
  • Test each allowed xx in the equation; the answer can be none.

Squeezed argument

u≥1 and u≤1 ⇒ u=1u\ge1\ \text{and}\ u\le1\ \Rightarrow\ u=1

Worked example

Solve cos⁡−1(x2−2x+2)+tan⁡−1(x−1)=0\cos^{-1}(x^2-2x+2)+\tan^{-1}(x-1)=0.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 31 January 2023 · Q68Moderate

Example 3 · Inverse Trigonometric Functions · Equations in Inverse Trigonometric Functions

Let y=f(x)y = f(x) represent a parabola with focus (−12,0)\left( -\frac{1}{2},0 \right) and directrix y=−12y = -\frac{1}{2}. Then S={x∈R:tan⁡−1(f(x))+sin⁡−1(f(x)+1)=π2}S =\left\{ x\in R:\tan^{- 1}(\sqrt{f(x)}) +\sin^{- 1}(\sqrt{f(x) + 1}) =\frac{\pi}{2} \right\} :

Find the domain first

Manipulating the equation before checking the domain wastes time and can give roots at which a term is undefined. When a square root and an inverse sine share an argument, find the allowed xx first.
Drill 2 more on when the domain pins x down

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)

  • Equations in inverse tangents

    Taking tangents

    tan⁡−1a+tan⁡−1b=c ⇒ a+b1−ab=tan⁡c\tan^{-1}a+\tan^{-1}b=c\ \Rightarrow\ \frac{a+b}{1-ab}=\tan c
  • Equations in inverse sines and cosines

    Isolate, then take the sine

    sin⁡−1u=θ ⇒ u=sin⁡θ,−π2≤θ≤π2\sin^{-1}u=\theta\ \Rightarrow\ u=\sin\theta,\quad -\tfrac{\pi}{2}\le\theta\le\tfrac{\pi}{2}
  • When the domain pins x down

    Squeezed argument

    u≥1 and u≤1 ⇒ u=1u\ge1\ \text{and}\ u\le1\ \Rightarrow\ u=1

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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