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

Question

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

Answer 1

Решение
cosa = 12/13    0 < a < π/2<br>tg(π/4 + a) = [tg(π/4) + tga] / [1 - tg(π/4)*tga] = (1 + tga) / (1 - tga)
sina = √(1 - cos²a) = √(1 - (12/13)²) = √(1 - 144/169) =
= √25/169 = 5/13
tga = sina/cosa
tga = 5/13 : 12/13 = 5/12
tg(π/4 + a) = (1 + 5/12) / (1 - 5/12) = 17/12 : 7/12 = 17/7 = 2 (3/7)