MathematicsMedium179×since 2002Q4293limx→π2(tan2x((2sin2x+3sinx+4)12−(sin2x+6sinx+2)12))\mathop {\lim }\limits_{x \to {\pi \over 2}} \left( {{{\tan }^2}x\left( {{{(2{{\sin }^2}x + 3\sin x + 4)}^{{1 \over 2}}} - {{({{\sin }^2}x + 6\sin x + 2)}^{{1 \over 2}}}} \right)} \right)x→2πlim(tan2x((2sin2x+3sinx+4)21−(sin2x+6sinx+2)21)) is equal toA112{1 \over {12}}121B−-−118{1 \over {18}}181C−-−112{1 \over {12}}121D16{1 \over {6}}61Check answerSkip