Resposta :

MATHSPHIS
[tex]cosx=\sqrt{1-sen^2x} \\ \\ cosx=\sqrt{1-(\frac{3}{5})^2} \\ \\ cosx=\sqrt{1-\frac{9}{25}} \\ \\ cosx=\sqrt{\frac{16}{25}}=\frac{4}{5}[/tex]

[tex]tgx=\frac{senx}{cosx}=\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4} \\ [/tex]

Sendo do 4o. quadrante tg x = -3/4

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