MathematicsHard66×since 2002Q4108Let x∗y=x2+y3x * y = {x^2} + {y^3}x∗y=x2+y3 and (x∗1)∗1=x∗(1∗1)(x * 1) * 1 = x * (1 * 1)(x∗1)∗1=x∗(1∗1). Then a value of 2sin−1(x4+x2−2x4+x2+2)2{\sin ^{ - 1}}\left( {{{{x^4} + {x^2} - 2} \over {{x^4} + {x^2} + 2}}} \right)2sin−1(x4+x2+2x4+x2−2) is :Aπ4{\pi \over 4}4πBπ3{\pi \over 3}3πCπ2{\pi \over 2}2πDπ6{\pi \over 6}6πCheck answerSkip