高一数学关于三角比1题，求助~急！

11 |

2 | 2009-8-6 14:05:47 | 显示全部楼层 |阅读模式

tan2αtan(30°-α)+tan2αtan(60°-α)+tan(30°-α)tan(60°-α)=tan2α[tan(30°-α)+tan(60°-α)]+tan(30°-α)tan(60°-α)=tan2α{tan(30°-α+60°-α)*[1-tan(30°-α)tan(60°-α)]}+tan(30°-α)tan(60°-α)=tan2αcot2α*[1-tan(30°-α)tan(60°-α)]}+tan(30°-α)tan(30°-α)=1...

使用道具举报