Tính tan20.tan40.tan60.tan80 mik đang cần gấp ạ,cảm ơn mn ạ

Đáp án:

 

Giải thích các bước giải:

Ta có `tan20.tan40.tan80`

`=(2sin20.sin40.sin80)/(2cos20.cos40.cos80)`

`=([cos(20-40)-cos(20+40)]sin80)/([cos(20+40)+cos(20-40)]cos80)`

`=((cos20-cos60).sin80)/((cos60+cos20)cos80)`

`=(2cos20sin80-2 . 1/2 . sin80)/(2 . 1/2 cos80+2cos20.cos80)` `(`vì `cos60=1/2)`

`=(sin(20+80)-sin(20-80)-sin80)/(cos80+cos(20+80)+cos(20-80))`

`=(sin100+sin60-sin80)/(cos80+cos100+cos60)`

`=(2cos((100+80)/2) sin((100-80)/2)+sqrt3/2)/(2cos((100+80)/2)cos((100-80)/2)+1/2)` `(`vì `sin60=sqrt3/2)`

`=(2cos90.sin10+sqrt3/2)/(2cos90.cos10+1/2)`

`=(sqrt3/2)/(1/2)` `(`vì `cos90=0)`

`=sqrt3=tan60.`

Do đó `tan20.tan40.tan60.tan80`

`=(tan20.tan40.tan80).tan60`

`=tan60 . tan60`

`=3.`