[cos40°+sin50°(1+√3tan10°)]/[sin70°√(1+cos40°)]=?

cos40°+sin50°×（1+√3tan10°）
=cos40°+sin50°×（tan60°-tan10°）/tan50°
=cos40°+（tan60°-tan10°）cos50°
=cos40°+√3cos50°-tan10°cos50°
=cos40°+√3sin40°-tan10°sin40°
=2[(1/2)cos40+(√3/2)sin40°]-(sin10°/cos10°)sin40°
=2(cos60°cos40°+sin60°sin40°)-[(sin10°)∧2/cos10°sin10°]sin40°
=2cos20°-[(1-cos20°)/sin20°]2sin20°cos20°
=2cos20°-2cos20°+2(cos20°) ∧2
=1+cos40°

sin70°√1+cos40°=sin70°(√2)cos20°=(√2)cos20°∧2
=√2/2（1+cos40°)

[cos40°+sin50°(1+√3tan10°)]/[sin70°√(1+cos40°)]=√2