JEE Main 202226 Jun, Morning shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002limx→12sin(cos−1x)−x1−tan(cos−1x)\mathop {\lim }\limits_{x \to {1 \over {\sqrt 2 }}} {{\sin ({{\cos }^{ - 1}}x) - x} \over {1 - \tan ({{\cos }^{ - 1}}x)}}x→21lim1−tan(cos−1x)sin(cos−1x)−x is equal to :AOption A. 2\sqrt 22BOption B. −2- \sqrt 2−2COption C. 12{1 \over {\sqrt 2 }}21DOption D. −12- {1 \over {\sqrt 2 }}−21CheckSkip · show solution
