If $(\sin^{-1}x)^2-(\cos^{-1}x)^2=a$ ; $0<x<1$, $a\neq 0$, then the value of $2x^2-1$ is :

Option 2: $\sin\left(\frac{2a}{\pi}\right)$

$\cos\left(\frac{2a}{\pi}\right)$

$\sin\left(\frac{2a}{\pi}\right)$

$\sin\left(\frac{4a}{\pi}\right)$

$\cos\left(\frac{4a}{\pi}\right)$

JEE Main 2021 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 27 August 2021, Shift 1, Maths Q76

Question 76 of 90 in this shift, Maths question 16 of 30, Section A.

(\sin^{-1}x)^2-(\cos^{-1}x)^2=a

;

0<x<1

a\neq 0

, then the value of

2x^2-1

is :

(1) $\cos\left(\frac{2a}{\pi}\right)$
(2) $\sin\left(\frac{2a}{\pi}\right)$ Official answer
(3) $\sin\left(\frac{4a}{\pi}\right)$
(4) $\cos\left(\frac{4a}{\pi}\right)$

Official answer

Option 2

NTA final key.

Chapter: Inverse Trigonometric Functions
Topic: Inverse Trig Identities
Idea tested: Complementary Identities

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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.