MathematicsMedium179×since 2002Q4284The 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 :A0B−-−12{1 \over 2}21C14{1 \over 4}41D−-−14{1 \over 4}41Check answerSkip