tan70=cot(90-70)=cot20=1/tan20
所以原式=(√3sin10+cos10)/tan20-2cos40
=2sin(10+arctan1/√3)/tan20-2cos40
=2sin40/tan20-2cos40
=2sin40*tan70-2cos40
=2sin40*cot20-2cos40
=4sin20cos20*cos20/sin20-2[2(cos20)^2-1]
=4(cos20)^2-4(cos20)^2+2
=2
tan70=cot(90-70)=cot20=1/tan20
所以原式=(√3sin10+cos10)/tan20-2cos40
=2sin(10+arctan1/√3)/tan20-2cos40
=2sin40/tan20-2cos40
=2sin40*tan70-2cos40
=2sin40*cot20-2cos40
=4sin20cos20*cos20/sin20-2[2(cos20)^2-1]
=4(cos20)^2-4(cos20)^2+2
=2