Pré-Vestibular ⇒ Trigonometria(soma de arcos)

[damaceno]

[tex3]
\mathsf{
\text{Se } \alpha, \beta, \alpha+\beta \notin \left\{ \frac{\pi}{2} + k\pi \right\},\; k \in \mathbb{Z}, \text{ então:} \\
\tg(\alpha+\beta) = \dfrac{\tg \alpha + \tg \beta}{1 - \tg \alpha \cdot \tg \beta} \\
\text{Dados: } a,b,c . \tg (b) = 2, \; \tg(a + b + c) = \dfrac{5}{4} \\
\text{Calcule: } \tg(a - b + c)
}
[/tex3]

Resposta

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-32. obs de alguma forma eu cheguei em 16/7

[damaceno]

[tex3]
\mathsf{
\text{Sabemos que } \tg b = 2 \quad \text{e} \quad \tg(a + b + c) = \dfrac{5}{4} \\

\tg(a + b + c) = \tg((a + c) + b) = \dfrac{\tg(a + c) + \tg b}{1 - \tg(a + c) \cdot \tg b} \\

\Rightarrow \dfrac{\tg(a + c) + 2}{1 - 2 \cdot \tg(a + c)} = \dfrac{5}{4} \\

\text{Multiplicando cruzado: } 5tg(a + c) + 10 = 4-8tg(a+c) \\\Rightarrow

13\tg(a + c) = -6\\
\Rightarrow \tg(a + c) = -\dfrac{6}{13} \\

\text{Queremos } \tg(a - b + c) = \tg((a + c) - b) \\

\tan(a-b+c) = \frac{-\frac{6}{13} - 2}{1 + (-\frac{6}{13} \cdot 2)} = \frac{-\frac{32}{13}}{\frac{1}{13}} = \boxed{-32}
}
[/tex3]