Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
If sin-1x2-cos-1x2=a; 0<x<1, a0, then the value of 2x2-1 is
MathematicsInverse Trigonometric FunctionsJEE MainJEE Main 2021 (27 Aug Shift 1)
Options:
  • A cos2aπ
  • B sin2aπ
  • C cos4aπ
  • D sin4aπ
Solution:
2310 Upvotes Verified Answer
The correct answer is: sin2aπ

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

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.