Question

已知tan[α/2]=2，则tanα的值为-[4/3]-[4/3]；[6sinα+cosα/3sinα−2cosα]的值

Answer 1

∵已知tan[α/2]=2，则tanα=

2tan

α

2

1−tan2

α

2=[4/1−4]=-[4/3]．

[6sinα+cosα/3sinα−2cosα]=[6tanα+1/3tanα−2]=[−8+1/−4−2]=[7/6]，

故答案为：-[4/3]；[7/6]．