Câu 12:

`A = (1 + sin(4α) - cos(4α)) / (1 + sin(4α) + cos(4α))`

`A = ((1 - cos(4α)) + sin(4α)) / ((1 + cos(4α)) + sin(4α))` 

`A = (2sin^2(2α) + 2sin(2α)cos(2α)) / (2cos^2(2α) + 2sin(2α)cos(2α))`

`A = (2sin(2α)(sin(2α) + cos(2α))) / (2cos(2α)(cos(2α) + sin(2α)))`

`A = sin(2α) / cos(2α)`

`A = tan(2α)`

`-> bbC`

Câu 13:

`A = (3 - 4cos(2α) + cos(4α)) / (3 + 4cos(2α) + cos(4α))` 

Ta có:

`cos(4α) = 2cos^2(2α) - 1`

Tử số: `3 - 4cos(2α) + (2cos^2(2α) - 1)`

`= 2cos^2(2α) - 4cos(2α) + 2`

`= 2(cos^2(2α) - 2cos(2α) + 1)`

`= 2(cos(2α) - 1)^2`

Mẫu số: `3 + 4cos(2α) + (2cos^2(2α) - 1)`

`= 2cos^2(2α) + 4cos(2α) + 2`

`= 2(cos^2(2α) + 2cos(2α) + 1)`

`= 2(cos(2α) + 1)^2`

`=> A = (2(cos(2α) - 1)^2) / (2(cos(2α) + 1)^2)`

`=> A = ((cos(2α) - 1) / (cos(2α) + 1))^2`

`=> A = ((-2sin^2(α)) / (2cos^2(α)))^2`

`=> A = (-sin^2(α) / cos^2(α))^2`

`=> A = (-tan^2(α))^2`

`A = tan^4(α)` 

`-> bbB`