Câu hỏi:

Ta có $\sin a = \dfrac{1}{3}$.
$\sin^2 a + \cos^2 a = 1$ suy ra $\cos^2 a = 1 - \dfrac{1}{9} = \dfrac{8}{9}$.
Do đó, $\cos a = \pm \dfrac{2\sqrt{2}}{3}$.
Trường hợp 1: $\cos a = \dfrac{2\sqrt{2}}{3}$. Khi đó, $\tan a = \dfrac{\sin a}{\cos a} = \dfrac{1}{2\sqrt{2}} = \dfrac{\sqrt{2}}{4}$ và $\cot a = \dfrac{1}{\tan a} = 2\sqrt{2}$.
$A = \dfrac{2\sqrt{2} - \dfrac{\sqrt{2}}{4}}{\dfrac{\sqrt{2}}{4} + 2(2\sqrt{2})} = \dfrac{\dfrac{7\sqrt{2}}{4}}{\dfrac{17\sqrt{2}}{4}} = \dfrac{7}{17}$.
Trường hợp 2: $\cos a = -\dfrac{2\sqrt{2}}{3}$. Khi đó, $\tan a = \dfrac{\sin a}{\cos a} = -\dfrac{1}{2\sqrt{2}} = -\dfrac{\sqrt{2}}{4}$ và $\cot a = \dfrac{1}{\tan a} = -2\sqrt{2}$.
$A = \dfrac{-2\sqrt{2} - (-\dfrac{\sqrt{2}}{4})}{-\dfrac{\sqrt{2}}{4} + 2(-2\sqrt{2})} = \dfrac{-\dfrac{7\sqrt{2}}{4}}{-\dfrac{17\sqrt{2}}{4}} = \dfrac{7}{17}$.
Vậy $A = \dfrac{7}{17}$.