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JEE Mains Maths · Inverse Trigonometric Functions

Sums of Inverse Tangents and Telescoping Series

Adding inverse tangents and cotangents with the addition formula, and summing series whose terms split into differences of two inverse tangents.

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Why this matters

Thirteen PYQs, ten of them multiple choice, and two from 2026. Four add two or three inverse tangents or cotangents with the addition formula; nine sum a series whose terms telescope. Two ideas cover the page.

Concept 1 of 2: The addition formula for inverse tangents

Taking the tangent of tan⁡−1a+tan⁡−1b\tan^{-1}a+\tan^{-1}b gives a+b1−ab\frac{a+b}{1-ab}. When ab<1ab<1, the sum lies in (−π2,π2)\left(-\frac\pi2,\frac\pi2\right) and equals tan⁡−1a+b1−ab\tan^{-1}\frac{a+b}{1-ab}. When a,b>0a,b>0 and ab>1ab>1, the sum is past π2\frac\pi2, so add π\pi. Convert inverse cotangents to inverse tangents first.

Definition

  • ab<1ab<1: tan⁡−1a+tan⁡−1b=tan⁡−1a+b1−ab\tan^{-1}a+\tan^{-1}b=\tan^{-1}\frac{a+b}{1-ab}.
  • a,b>0, ab>1a,b>0,\ ab>1: add π\pi. a,b<0, ab>1a,b<0,\ ab>1: subtract π\pi.
  • ab>−1ab>-1: tan⁡−1a−tan⁡−1b=tan⁡−1a−b1+ab\tan^{-1}a-\tan^{-1}b=\tan^{-1}\frac{a-b}{1+ab}.
  • cot⁡−1x=tan⁡−11x\cot^{-1}x=\tan^{-1}\frac1x for x>0x>0, and π+tan⁡−11x\pi+\tan^{-1}\frac1x for x<0x<0.

Addition formula

tan⁡−1a+tan⁡−1b=tan⁡−1a+b1−ab(ab<1)\tan^{-1}a+\tan^{-1}b=\tan^{-1}\frac{a+b}{1-ab}\quad(ab<1)

Worked example

Find tan⁡−12+tan⁡−13\tan^{-1}2+\tan^{-1}3.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 6 Apr 2026 Shift 1 · Q63Moderate

Example 1 · Inverse Trigonometric Functions · Sums of Inverse Tangents and Telescoping Series

Let 0<α<1,β=13α0 <\alpha< 1,\beta=\frac{1}{3\alpha} and tan⁡−1(1−α)+tan⁡−1(1−β)=π4\tan^{- 1}(1 -\alpha) +\tan^{- 1}(1 -\beta) =\frac{\pi}{4}. Then 6(α+β)6(\alpha+\beta) is equal to:

When the product exceeds 1

For positive a,ba,b with ab>1ab>1, a+b1−ab\frac{a+b}{1-ab} is negative although both angles are positive. The formula alone gives a negative angle; add π\pi.
Drill 3 more on the addition formula for inverse tangents

Concept 2 of 2: Telescoping sums of inverse tangents

Read the subtraction formula backwards: tan⁡−1b−a1+ab=tan⁡−1b−tan⁡−1a\tan^{-1}\frac{b-a}{1+ab}=\tan^{-1}b-\tan^{-1}a. If each term of a series can be written this way with bb of one term equal to aa of the next, the sum collapses to the last bb minus the first aa. The work is spotting aa and bb: write the denominator as 1+ab1+ab where b−ab-a is the numerator.

Definition

  • tan⁡−1b−a1+ab=tan⁡−1b−tan⁡−1a\tan^{-1}\frac{b-a}{1+ab}=\tan^{-1}b-\tan^{-1}a for ab>−1ab>-1.
  • 11+n(n+1)\frac1{1+n(n+1)}: a=na=n, b=n+1b=n+1.
  • A cotangent term cot⁡−1c\cot^{-1}c with c>0c>0 is tan⁡−11c\tan^{-1}\frac1c; scale numerator and denominator until the numerator is b−ab-a.
  • ∑r=1n(tan⁡−1br−tan⁡−1br−1)=tan⁡−1bn−tan⁡−1b0\sum_{r=1}^{n}\left(\tan^{-1}b_r-\tan^{-1}b_{r-1}\right)=\tan^{-1}b_n-\tan^{-1}b_0, and tan⁡−1bn→π2\tan^{-1}b_n\to\frac\pi2 as bn→∞b_n\to\infty.

Telescoping term

tan⁡−1b−a1+ab=tan⁡−1b−tan⁡−1a\tan^{-1}\frac{b-a}{1+ab}=\tan^{-1}b-\tan^{-1}a

Worked example

Find ∑r=1∞tan⁡−12rr4+r2+2\sum_{r=1}^{\infty}\tan^{-1}\frac{2r}{r^4+r^2+2}.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 5 Apr 2026 Shift 1 · Q74Moderate

Example 2 · Inverse Trigonometric Functions · Sums of Inverse Tangents and Telescoping Series

If π4+∑p=111tan⁡−1(2p−11+22p−1)=α\frac{\pi}{4} + \sum_{p = 1}^{11} \tan^{- 1}\left( \frac{2^{p - 1}}{1 + 2^{2p - 1}} \right) = \alpha, then tan⁡α\tan\alpha is equal to ____\_\_\_\_ .

Keep the order of the difference

The term must be tan⁡−1(next)−tan⁡−1(this)\tan^{-1}(\text{next})-\tan^{-1}(\text{this}), so the numerator is the larger minus the smaller. With the order reversed the sum comes out with the wrong sign, and it will not match the options.
Drill 8 more on telescoping sums of inverse tangents

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 (2)

  • The addition formula for inverse tangents

    Addition formula

    tan⁡−1a+tan⁡−1b=tan⁡−1a+b1−ab(ab<1)\tan^{-1}a+\tan^{-1}b=\tan^{-1}\frac{a+b}{1-ab}\quad(ab<1)
  • Telescoping sums of inverse tangents

    Telescoping term

    tan⁡−1b−a1+ab=tan⁡−1b−tan⁡−1a\tan^{-1}\frac{b-a}{1+ab}=\tan^{-1}b-\tan^{-1}a

Watch out for (2)

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.

Take the Inverse Trigonometric Functions chapter testDrill all 13 questions