MathematicsMedium179×since 2002Q4270limx→0 (1−cos2x)22x tanx −xtan2x\mathop {\lim }\limits_{x \to 0} \,{{{{\left( {1 - \cos 2x} \right)}^2}} \over {2x\,\tan x\, - x\tan 2x}}x→0lim2xtanx−xtan2x(1−cos2x)2 is :A−-− 2B−-− 12{1 \over 2}21C12{1 \over 2}21D2Check answerSkip