Question

求值：(1+cos20°）/2sin20°-sin10°(1/tan5°-tan5°)

Answer 1

[(1+cos20°）/2sin20°]-sin10°(1/tan5°-tan5°)
=[(1+cos20°)/2sin20°]-sin10°(cot5°-tan5°)
=[(1+cos20°)/4sin10°cos10°]-sin10°(cot5°-tan5°)
=(2cos10°/4sin10°)-2sin5°cos5°(cot5°-tan5°)
=(cos10°/2sin10°)-2((cos5°)＾2-(sin5°)＾2)
=(cos10°/2sin10°)-2cos10°
=(cos10°-4sin10°cos10°)/2sin10°
=(sin80°-2sin20°)/2sin10°
=((sin80°-sin20°)-sin20°)/2sin10°
=(2cos50°sin30°-sin20°)/2sin10°
=(sin40°-sin20°)/2sin10°
={2cos30°sin10°)/2sin10°
=cos30°
=（根号3）/2

Answer 2

(1+cos20°）/2sin20°-sin10°(1/tan5°-tan5°)
=(2cos²10°)/(4sin10°cos10°)-sin10°(1-tan²5°)/tan5°
=cos10°/2sin10°-sin10°X2/tan10°
=cos10°/2sin10°-2cos10°
=1/（2tan10°)-2cos10°