JEE Main 202117 Mar, Evening shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002The value of the limit limθ→0tan(πcos2θ)sin(2πsin2θ)\mathop {\lim }\limits_{\theta \to 0} {{\tan (\pi {{\cos }^2}\theta )} \over {\sin (2\pi {{\sin }^2}\theta )}}θ→0limsin(2πsin2θ)tan(πcos2θ) is equal to :AOption A. 0BOption B. −-−12{1 \over 2}21COption C. 14{1 \over 4}41DOption D. −-−14{1 \over 4}41CheckSkip · show solution