Question

6sin150°cos150°/cos^2 30°- sin30°
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Answer 1

$\frac{6sin150^0cos150^0}{cos^230^0-sin30^0}= \frac{6sin(180^0-30^0)cos(180^0-30^0)}{ (\frac{ \sqrt{3} }{2})^2 - \frac{1}{2} } = \frac{6* sin30^0*(-cos30^0) }{ \frac{3}{4} - \frac{2}{4} } =$$\frac{6* \frac{1}{2}* \frac{- \sqrt{3} }{2} }{ \frac{1}{4} }=- \frac{6 \sqrt{3} *4}{1*4}=-6 \sqrt{3}$