$sin^2∝+cos^2∝=1$
$⇔\dfrac{4}{9}+cos^2∝=1$
$⇒cos^2∝=\dfrac{5}{9}$
$⇒cos∝=\dfrac{\sqrt{5}}{3}$
$tan∝=\dfrac{sin∝}{cos∝}=\dfrac{2}{3}:\dfrac{\sqrt{5}}{3}=\dfrac{2}{\sqrt{5}}=\dfrac{2\sqrt{5}}{5}$
$⇒cot∝=\dfrac{1}{tan∝}=\dfrac{\sqrt{5}}{2}$
$A=tan∝-\dfrac{cos∝}{cot∝}$
$=\dfrac{2\sqrt{5}}{5}-\dfrac{\sqrt{5}}{3} : \dfrac{\sqrt{5}}{2}$
$=\dfrac{2\sqrt{5}}{5}-\dfrac{\sqrt{5}}{3} × \dfrac{2}{\sqrt{5}}$
$=\dfrac{2\sqrt{5}}{5}-\dfrac{2}{3}$
$=\dfrac{6\sqrt{5}}{15}-\dfrac{10}{15}$
$=\dfrac{6\sqrt{5}-10}{15}$
Vậy $A=\dfrac{6\sqrt{5}-10}{15}$
