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JEE Main 2022 · MathsMultiple choiceSingle correctEasyCalculation

JEE Main 25 June 2022, Shift 2, Maths Q19

Question 19 of 90 in this shift, Maths question 19 of 30, Section A.

The value of

\tan^{-1}\left(\frac{\cos\left(\frac{15\pi}{4}\right) - 1}{\sin\left(\frac{\pi}{4}\right)}\right)

is equal to :

(1) $-\frac{\pi}{4}$
(2) $-\frac{\pi}{8}$ Official answer
(3) $-\frac{5\pi}{12}$
(4) $-\frac{4\pi}{9}$

Official answer

Option 2

NTA final key.

Chapter: Inverse Trigonometric Functions
Topic: Inverse Trig Identities
Idea tested: Periodic Reduction of sin-1(sin x) and cos-1(cos x)

Same idea in other shifts

Asked 2× in all

31 Jan 2024, Shift 2 · Q20If $a=\sin^{-1}(\sin(5))$ and $b=\cos^{-1}(\cos(5))$ , then $a^2+b^2$ is equal toMediumSingle 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.