JEE Main 202127 Aug, Morning shiftMathematicsEasyTrigonometry118× since 2002If (sin−1x)2−(cos−1x)2=a{({\sin ^{ - 1}}x)^2} - {({\cos ^{ - 1}}x)^2} = a(sin−1x)2−(cos−1x)2=a; 0 < x < 1, a ≠\ne= 0, then the value of 2x² −-− 1 is :AOption A. cos(4aπ)\cos \left( {{{4a} \over \pi }} \right)cos(π4a)BOption B. sin(2aπ)\sin \left( {{{2a} \over \pi }} \right)sin(π2a)COption C. cos(2aπ)\cos \left( {{{2a} \over \pi }} \right)cos(π2a)DOption D. sin(4aπ)\sin \left( {{{4a} \over \pi }} \right)sin(π4a)CheckSkip · show solution
