JEE Main 2024 · MathsMultiple choiceSingle correctMediumApplication

JEE Main 1 February 2024, Shift 1, Maths Q20

Question 20 of 90 in this shift, Maths question 20 of 30, Section A.

\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}

0<A,B,C<\frac{\pi}{2}

, then

A+B

is equal to :

(1) $C$ Official answer
(2) $\pi-C$
(3) $2\pi-C$
(4) $\frac{\pi}{2}-C$

Official answer

Option 1

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
26 Feb 2021, Shift 2 · Q64If $0 < a, b < 1$ , and $\tan^{-1}a + \tan^{-1}b = \frac{\pi}{4}$ , then the value of…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.