Question

se x pertence ao quadrante, e sen x =3/5 Então sen x.tg x e igual a ?

Answer 1

$sen^2x+cos^2x=1\\cos^2x=1-sen^2x\\cosx=\sqrt{1-sen^2x}\\cosx=\sqrt{1-(\frac{3}{5})^2}=\frac{4}{5}\\tgx=\frac{senx}{cosx}=\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}\\\\senx.tgx=\frac{3}{5}.\frac{3}{4}=\frac{9}{20}$