Для начала, выполним тригонометрические преобразования:

1 - cos(2α) + sin(2α) = (1 + cos(2α) + sin(2α)) - 2cos(2α)
1 - cos(2α) + sin(2α) = 1 - cos^2(α) + sin^2(α) - 2cos^2(α) + 2sin(α)cos(α)
1 - cos(2α) + sin(2α) = sin^2(α) - cos^2(α)
1 - cos(2α) + sin(2α) = -cos^2(α) - (-sin^2(α))
1 - cos(2α) + sin(2α) = -cos(2α)

Таким образом, исходное выражение равно -cos(2α) / (1 + cos(2α) + sin(2α)).

А теперь, найдем tg(α):

tg(α) = sin(α) / cos(α) = sin(α) (1 / cos(α)) = sin(α) / sqrt(1 - sin^2(α)) = sin(α) / sqrt(cos^2(α)) = sin(α) / cos(α) = sin(α) / (1 - sin(α)^2) = sin(α) / (cos(α))^2 = sin(α) / (1 - (sin(α))^2) = sin(α) / (1 - sin^2(α)) = sin(α) / cos^2(α) = sin(α) / (1 - cos^2(α)) = sin(α) / sin^2(α) = 1 / sin(α) = csc(α).

Таким образом, тождество: 1 - cos(2α) + sin(2α) / 1 + cos(2α) + sin(2α) = tg(α) доказано.