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JEE Main 2025 · MathsMultiple choiceSingle correctHardMulti-step

JEE Main 8 April 2025, Shift 2, Maths Q13

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

The value of

\cot^{-1}\left(\frac{\sqrt{1+\tan^2(2)}-1}{\tan(2)}\right) - \cot^{-1}\left(\frac{\sqrt{1+\tan^2\left(\frac{1}{2}\right)}+1}{\tan\left(\frac{1}{2}\right)}\right)

is equal to

(1) $\pi - \frac{5}{4}$ Official answer
(2) $\pi - \frac{3}{2}$
(3) $\pi + \frac{5}{2}$
(4) $\pi + \frac{3}{2}$

Official answer

Option 1

NTA final key.

Chapter: Inverse Trigonometric Functions
Topic: Inverse Trig Identities
Idea tested: Inverse Trig Simplification via tan(θ)

Same idea in other shifts

Asked 2× in all

24 Jun 2022, Shift 2 · Q1Let $x*y = x^2 + y^3$ and $(x*1)*1 = x*(1*1)$ . Then a value of $2\sin^{-1}\left(\frac{x^4+x^2-2}{x^4+x^2+2}\right)$ isMediumSingle 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.