Question

Sabendo que:
sen²x + cos²x=1

Determine tg x e sec x, sendo que sen x= -1/4

Answer 1

Resposta:

Explicação passo-a-passo:

$sen^x+cos^2x=1\\(\frac{-1}{4})^2+cos^2x=1\\\\\frac{1}{16} +cos^2x=1\\cos^2x=1-\frac{1}{16}\\cosx= \sqrt{\frac{15}{16}}\\cosx= \frac{\sqrt{15}}{4}}$

$tg x = \frac{senx}{cosx} = \frac{\frac{-1}{4}}{\frac{\sqrt{15}}{4}}= \frac{-1}{\sqrt{15}}= \frac{-\sqrt{15}}{15}$

$secx= \frac{1}{cosx} = \frac{1}{\frac{\sqrt{15}}{4}}= \frac{4}{\sqrt{15}}= \frac{4\sqrt{15}}{15}}$