$tan∝=\dfrac{1}{cot∝}=\dfrac{4}{3}$
$tan∝=\dfrac{sin∝}{cos∝}=\dfrac{4}{3}$
$⇒sin∝=\dfrac{4}{3}cos∝$
$sin^2∝+cos^2∝=1$
$⇔\left ( \dfrac{4}{3}cos∝ \right )^2+cos^2∝=1$
$⇔\dfrac{16}{9}cos^2∝+cos^2∝=1$
$⇔\dfrac{25}{9}cos^2∝=1$
$⇔cos∝=\dfrac{3}{5}$
$⇒sin∝=\dfrac{4}{5}$
Vậy $sin∝=\dfrac{4}{5} ; cos∝=\dfrac{3}{5} ; tan∝=\dfrac{4}{3}$