JEE Main 2026 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 6 April 2026, Shift 1, Maths Q13

Question 13 of 75 in this shift, Maths question 13 of 25, Section A.

Let

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:

(1)6
(2)7Official answer
(3)8
(4)9

Official answer

Option 2

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

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
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.