Verified Solution • 2025–26 Session moderate

QJEEM21Mad62f: JEE Main 2021 Mathematics - Class 12 JEEM21Mad62f Solution: If ; 0 < x < 1, a 0, then the value of 2x 2 1 is :

If (sin⁡−1x)2−(cos⁡−1x)2=a{({\sin ^{ - 1}}x)^2} - {({\cos ^{ - 1}}x)^2} = a ; 0 < x < 1, a ≠\ne 0, then the value of 2x2 −- 1 is :

Multiple Choice Options

  • A cos⁡(4aπ)\cos \left( {{{4a} \over \pi }} \right)
  • B sin⁡(2aπ)\sin \left( {{{2a} \over \pi }} \right)
  • C cos⁡(2aπ)\cos \left( {{{2a} \over \pi }} \right)
  • D sin⁡(4aπ)\sin \left( {{{4a} \over \pi }} \right)

Step-by-Step Verified Solution

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

=(sin⁡−1x+cos⁡−1x)(sin⁡−1x−cos⁡−1x)= ({\sin ^{ - 1}}x + {\cos ^{ - 1}}x)({\sin ^{ - 1}}x - {\cos ^{ - 1}}x)

=π2(π2−2cos⁡−1x)= {\pi \over 2}\left( {{\pi \over 2} - 2{{\cos }^{ - 1}}x} \right)

⇒2cos⁡−1x=π2−2aπ\Rightarrow 2{\cos ^{ - 1}}x = {\pi \over 2} - {{2a} \over \pi }

⇒cos⁡−1(2x2−1)=π2−2aπ\Rightarrow {\cos ^{ - 1}}(2{x^2} - 1) = {\pi \over 2} - {{2a} \over \pi }

⇒2x2−1=cos⁡(π2−2aπ)\Rightarrow 2{x^2} - 1 = \cos \left( {{\pi \over 2} - {{2a} \over \pi }} \right)

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