[(1+cos20°)/(2sin20°)]-sin10°(cot5°-tan5°)
2个回答

我们一个个来算,

(1+cos20°)/(2sin20°)=2cos²10°/(4sin10°cos10°)=cos10°/(2sin10°),

而cot5°-tan5°

=cos5°/sin5°-sin5°/cos5°

=(cos²5°-sin²5°)/(sin5°cos5°)

=cos10°/(1/2sin10°)

=2cos10°/sin10°,

所以原式=cos10°/(2sin10°)-2cos10°

= (cos10°-2sin20°)/(2sin10°)

=[cos10°-2sin(30°-10°)]/(2sin10°)

=(2cos30°sin10°))/(2sin10°)

=cos30°

=√3/2.