JEE Main 2005Annual, Paper 1MathematicsMediumTrigonometry118× since 2002If cos−1x−cos−1y2=α,{\cos ^{ - 1}}x - {\cos ^{ - 1}}{y \over 2} = \alpha ,cos−1x−cos−12y=α, then 4x2−4xycosα+y24{x^2} - 4xy\cos \alpha + {y^2}4x2−4xycosα+y2 is equal to :AOption A. 2sin2α2\sin 2\alpha2sin2αBOption B. 444COption C. 4sin2α4{\sin ^2}\alpha4sin2αDOption D. −4sin2α-4{\sin ^2}\alpha−4sin2αCheckSkip · show solution
