JEE Main 2021 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 26 February 2021, Shift 2, Maths Q64

Question 64 of 90 in this shift, Maths question 4 of 30, Section A.

0 < a, b < 1

, and

\tan^{-1}a + \tan^{-1}b = \frac{\pi}{4}

, then the value of

(a+b) - \left(\frac{a^2+b^2}{2}\right) + \left(\frac{a^3+b^3}{3}\right) - \left(\frac{a^4+b^4}{4}\right) + \ldots

is :

(1) $e$
(2) $e^2-1$
(3) $\log_e 2$ Official answer
(4) $\log_e\left(\frac{e}{2}\right)$

Official answer

Option 3

NTA final key.

Chapter: Inverse Trigonometric Functions
Topic: Inverse Trig Identities
Idea tested: Tan-1 Sum and Difference

Same idea in other shifts

Asked 4× in all

6 Apr 2026, Shift 1 · Q13Let $0<\alpha<1$ , $\beta=\frac{1}{3\alpha}$ and $\tan^{-1}(1-\alpha)+\tan^{-1}(1-\beta)=\frac{\pi}{4}$ . Then $6(\alpha+\beta)$ is equal to:MediumSingle correct
1 Feb 2024, Shift 1 · Q20If $\tan A=\frac{1}{\sqrt{x(x^2+x+1)}}$ , $\tan B=\frac{\sqrt{x}}{\sqrt{x^2+x+1}}$ and $\tan C=\left(x^{-3}+x^{-2}+x^{-1}\right)^{1/2}$ ,…MediumSingle correct
17 Mar 2021, Shift 1 · Q80The sum of possible values of $x$ for $\tan^{-1}(x+1)+\cot^{-1}\left(\frac{1}{x-1}\right)=\tan^{-1}\left(\frac{8}{31}\right)$ is :HardSingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.