Зная, что cos a=
\frac{12}{13}
0 < a <п/2
Найдите tg( \frac{ \pi }{4} } +a)

Решениеtg(π/4 + a) = [tg(π/) + tga] / [1 - tg(π/4)tga] == (1 + tga) / (1 - tga)cosa = 12/13sina = √(1 - cos²a) = √(1 - (12/13)²) = √(1 - 144/169) == √25/169) = 5/13tga = sina/cosa = 5/13 : 12/13 = 5/12tg(π/4 + a) = [1 + (5/12)] / [1 - (5/12)] = = 17/12 : 7/12 = 17/7 = 2 (3/7)