Question

URGENTE
Sabendo que cos x = -5/13 e π/2 < × < π , calcule o valor de tg x.

Answer 1

Olá Gabriela!

 Problemas desse tipo são solucionados fazendo uso da fórmula: $\mathsf{\sin^2 x + \cos^2 x = 1}$.

 Entretanto, devemos fazer uso do intervalo dado no enunciado!

 Segue,

$\\ \mathsf{\sin^2 x + \cos^2 x = 1} \\\\ \mathsf{\sin^2 x + \left ( - \frac{5}{13} \right )^2 = 1} \\\\ \mathsf{\sin^2 x + \frac{25}{169} = 1} \\\\ \mathsf{\sin^2 x = 1 - \frac{25}{169}} \\\\ \mathsf{\sin^2 x = \frac{144}{169}} \\\\ \boxed{\sin x = \pm \frac{12}{13}}$

 Ora, de acordo com o intervalo - 2º quadrante, tiramos que $\boxed{\mathsf{\sin x = + \frac{12}{13}}}$.

 Com efeito,

$\\ \mathsf{\tan x = \frac{\sin x}{\cos x}} \\\\\\ \mathsf{\tan x = \frac{\frac{12}{13}}{\frac{- 5}{13}}} \\\\\\ \mathsf{\tan x = \frac{12}{13} \div \frac{- 5}{13}} \\\\\\ \mathsf{\tan x = \frac{12}{13} \times \frac{13}{- 5}} \\\\\\ \boxed{\boxed{\mathsf{\tan x = - \frac{12}{5}}}}$