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Malumtt
@Malumtt
December 2019
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Determine o valor da tg x, sabendo que sec x= √5 e 3π/2 <x<2π
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albertrieben
Boa noite Malu
sec(x) =
√5
cos(x) = 1/
√5 =
√5/5
cos
²(x) + sen²(x) = 1
1/5 + sen²(x) = 5/5
sen²(x) = 5/5 - 1/5 = 4/5
no 4° quadrante
3π/2 <x<2π o sen . negativo
sen(x) = -2√5/5
tg(x) = sen(x)/cos(x) = (-2√5/5)/(√5/5)
tg(x) = -2
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sec(x) = √5
cos(x) = 1/√5 = √5/5
cos²(x) + sen²(x) = 1
1/5 + sen²(x) = 5/5
sen²(x) = 5/5 - 1/5 = 4/5
no 4° quadrante 3π/2 <x<2π o sen . negativo
sen(x) = -2√5/5
tg(x) = sen(x)/cos(x) = (-2√5/5)/(√5/5)
tg(x) = -2