求1/sin50'+根号下3/cos50'的值?
1个回答

（sin50°(1 √3*tan10°)-cos20°）/（cos80°*√（1-cos20°）) =（sin50°(1 √3*sin10°/cos10°)-cos20°）/（cos80°*√（1-cos20°）） =(sin50°((cos10° √3*sin10°)/cos10°)-cos20°）/（cos80°*√(2(sin10°)^2)) =(sin50°((1/2)cos10° (√3/2)*sin10°)/((1/2)*cos10°)-cos20°）/（cos80°*√(2(sin10°)^2)) =(sin50°(sin30°*cos10° cos30°*sin10°)/(1/2)*cos10°-cos20°）/（(√2)cos80°*sin10°) =(sin50°*sin40°/(1/2)*cos10°-cos20°）/（(√2)sin10°*sin10°) =(sin50°*cos50°/(1/2)*cos10°-cos20°）/（(√2)sin10°*sin10°) =((1/2)sin100°/(1/2)*cos10°-cos20°）/（(√2)sin10°*sin10°) =((1/2)sin80°/(1/2)*cos10°-cos20°）/（(√2)sin10°*sin10°) =((1/2)cos10/(1/2)*cos10°-cos20°）/（(√2)sin10°sin10°) =(1-cos20°)/（(√2)sin10°sin10°) =2(sin20°)^2/(√2)(sin20°)^2 =2/√2 =√2

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