一道数学题,最好写下过程：sin40'（tan10'减根号3）＜'表示单位度＞

sin40'*[tan10'-3^(1/2)]
=sin40'*(tan10'-tan60')
=sin40'*(sin10'cos60'-sin60'cos10')/(cos10'cos60')
=-sin40'sin50'/(cos10'cos60')
=-2sin40'cos40'/cos10'
=-sin80'/sin80'
=-1