Question

se x está n 3º quadrante e sen(x)= -12/13, calcular o valor do cos (x), tg(x), sec(x), cossec(x) e cotg(x)

Answer 1

  cosx^2 = 1-(-12/13)^2
  cosx^2 = 1 -144/169
  cosx^2 = (169-144)/169
  cosx^2 = 25/169

   cosx^ = - 5/13

   tgx = senx  = - 12/13   ==> tgx = 12/5
            cox        - 5/13

cotgx = cosx  =  -5/13   ==>  cotgx = 5/13
             senx     -12/13

cosec = 1/senx  =  - 13/12

secx = 1/cox = - 13/5

Answer 2

$\text{Se} \ \text{sen} \ x = -\frac{12}{13}: \\\\ \bullet \text{cossec} \ x = -\frac{13}{12} \text{, pois cossec} \ x = \text{(sen} \ x)^{-1} \\\\ \bullet \text{cos} \ x: \\\\ \circ \text{hipotenusa = 13 (denominador de sen} \ x) \\ \circ \text{cateto oposto = 12 (numerador de sen} \ x) \\ \circ \text{cateto adjacente:} \\\\ 13^2 = 12^2 + {c_a}^{2} \\ 169 = 144 + {c_a}^{2} \\ {c_a}^{2} = 25 \\ c_a = 5 \\\\ \text{cos} \ x = \frac{\text{cateto adjacente}}{\text{hipotenusa}}$

$\text{cos} \ x = -\frac{5}{13} \ (\text{cos} \ x < 0 \ \text{pois} \ x \in 3^\circ \ \text{quadrante}) \\\\ \bullet \text{tg} \ x = \frac{\text{sen} \ x}{\text{cos} \ x} \\\\ \text{tg} \ x = \frac{-\frac{12}{13}}{-\frac{5}{13}} = -\frac{12}{13} \cdot (-\frac{13}{5}) = \frac{156}{65} = \frac{12}{5} \\\\ \bullet \text{cotg} \ x = \frac{5}{12}, \text{pois cotg} \ x = (\text{tg} \ x)^{-1}$