JEE Main 2021MathematicsInverse Trigonometric FunctionsMediumMCQ

JEE Main 2021 — Inverse Trigonometric Functions Question with Solution

JEE Main 2021 (27 Aug Shift 1)

Question

If sin-1x2-cos-1x2=a; 0<x<1, a≠0, then the value of 2x2-1 is

Choose an option

Show full solutionCorrect option: B
Correct answer
Bsin2aπ

Step-by-step explanation

sin-1x2-cos-1x2=a

Let sin-1x=t

x=sin t

sin-1sin t2-cos-1sin t2=a

t2-π2-t2=a

t2-π24+t2-πt=a

πt-π24=a

t=aπ+π4

x=sinaπ+π4

2x2-1=2sin2aπ+π4-1

=2sinaπ cosπ4+cosaπ sinπ42-1

=sinaπ+cosaπ2-1

=sin2aπ+cos2aπ+2cosaπsinaπ-1

=2cosaπsinaπ

=sin2aπ

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About this question

This is a previous-year question from JEE Main 2021, covering the Inverse Trigonometric Functions chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.