求(1+cos20°)/cos10°-sin10°(1/tan5°-tan5°)的值

网友回答

(1+cos20°)/cos10°-sin10°(1/tan5°-tan5°)
=(1+2cos^2(10)-1)/cos10-sin10[cos5/sin5-sin5/cos5]
=2cos10-sin10(cos^2(5)-sin^2(5))/(sin5cos5)
=2cos10-sin10(cos10/(1/2sin10)
=2cos10-2cos10
=0======以下答案可供参考======
供参考答案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
供参考答案2:
(1+cos20°)/cos10°-sin10°(1/tan5°-tan5°)
=(1+cos20°)/cos10°-2sin5°cos5°（cos5°/sin5°-sin5°/cos5°）
=(1+cos20°)/cos10°-(2cos5°cos5°-2sin5°sin5°)
=0