MathematicsHard66×since 2002Q4076Let α=tan(5π16sin(2cos−1(15)))\alpha = \tan \left( {{{5\pi } \over {16}}\sin \left( {2{{\cos }^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)} \right)} \right)α=tan(165πsin(2cos−1(51))) and β=cos(sin−1(45)+sec−1(53))\beta = \cos \left( {{{\sin }^{ - 1}}\left( {{4 \over 5}} \right) + {{\sec }^{ - 1}}\left( {{5 \over 3}} \right)} \right)β=cos(sin−1(54)+sec−1(35)) where the inverse trigonometric functions take principal values. Then, the equation whose roots are α\alphaα and β\betaβ is :A15x2−8x−7=015{x^2} - 8x - 7 = 015x2−8x−7=0B5x2−12x+7=05{x^2} - 12x + 7 = 05x2−12x+7=0C25x2−18x−7=025{x^2} - 18x - 7 = 025x2−18x−7=0D25x2−32x+7=025{x^2} - 32x + 7 = 025x2−32x+7=0Check answerSkip