JEE Main 2004Annual, Paper 1MathematicsHardTrigonometry118× since 2002If u=a2cos2θ+b2sin2θ+a2sin2θ+b2cos2θu = \sqrt {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } + \sqrt {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta }u=a2cos2θ+b2sin2θ+a2sin2θ+b2cos2θ then the difference between the maximum and minimum values of u2{u^2}u2 is given by :AOption A. (a−b)2{\left( {a - b} \right)^2}(a−b)2BOption B. 2a2+b22\sqrt {{a^2} + {b^2}}2a2+b2COption C. (a+b)2{\left( {a + b} \right)^2}(a+b)2DOption D. 2(a2+b2)2\left( {{a^2} + {b^2}} \right)2(a2+b2)CheckSkip · show solution
