若x+y+z=nπ,求证:tanx+tany+tanz=tanxtanytanz
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/19 01:34:55
![若x+y+z=nπ,求证:tanx+tany+tanz=tanxtanytanz](/uploads/image/z/12514369-49-9.jpg?t=%E8%8B%A5x%2By%2Bz%3Dn%CF%80%2C%E6%B1%82%E8%AF%81%3Atanx%2Btany%2Btanz%3Dtanxtanytanz)
若x+y+z=nπ,求证:tanx+tany+tanz=tanxtanytanz
若x+y+z=nπ,求证:tanx+tany+tanz=tanxtanytanz
若x+y+z=nπ,求证:tanx+tany+tanz=tanxtanytanz
tan(x+y)=(tanx+tany)/(1-tanxtany)
所以 tanx+tany=tan(x+y)(1-tanxtany)
x+y+z=nπ
所以 tanz=tan[nπ-(x+y)]=-tan(x+y)
所以 tanx+tany+tanz=tan(x+y)(1-tanxtany)-tan(x+y)
=-tan(x+y)tanxtany
=tanxtanytanz
得证