三角恒等变换综合求值:cos40°+sin50°(1+根号3tan10°)/sin70°根号(1+cos40°)
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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