Зная, что cos a=
\frac{12}{13}
0 < a <п/2
Найдите tg( \frac{ \pi }{4} } +a)
\frac{12}{13}
0 < a <п/2
Найдите tg( \frac{ \pi }{4} } +a)
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tg(π/4 + a) = [tg(π/) + tga] / [1 - tg(π/4)tga] =
= (1 + tga) / (1 - tga)
cosa = 12/13
sina = √(1 - cos²a) = √(1 - (12/13)²) = √(1 - 144/169) =
= √25/169) = 5/13
tga = sina/cosa = 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)