JEE Main 201910 Apr, Evening shiftMathematicsMediumTrigonometry118× since 2002If cos−1x−cos−1y2=α{\cos ^{ - 1}}x - {\cos ^{ - 1}}{y \over 2} = \alphacos−1x−cos−12y=α,where –1 ≤\le≤ x ≤\le≤ 1, – 2 ≤\le≤ y ≤\le≤ 2, x ≤\le≤ y2{y \over 2}2y , then for all x, y, 4x² – 4xy cos α\alphaα + y² is equal to :AOption A. 4 sin² α\alphaαBOption B. 2 sin² α\alphaαCOption C. 4 sin² α\alphaα - 2x²y²DOption D. 4 cos² α\alphaα + 2x²y²CheckSkip · show solution
