Đáp án:
$\begin{array}{l}
\tan x = \dfrac{1}{3} = \dfrac{{\sin x}}{{\cos x}}\\
\Rightarrow \cos x = 3\sin x\\
A = \dfrac{{\sin 3x - \cos 3x}}{{\cos 3x}}\\
= \dfrac{{\sin 3x}}{{\cos 3x}} - 1\\
= \dfrac{{3{\mathop{\rm sinx}\nolimits} - 4{{\sin }^3}x}}{{4{{\cos }^3}x - 3\cos x}} - 1\\
= \dfrac{{3\sin x - 4{{\sin }^3}x}}{{4.27{{\sin }^3}x - 3.3\sin x}} - 1\\
= \dfrac{{3 - 4{{\sin }^2}x}}{{108{{\sin }^2}x - 9}} - 1\\
Do:\dfrac{1}{{{{\cos }^2}x}} = {\tan ^2}x + 1 = \dfrac{{10}}{9}\\
\Rightarrow {\cos ^2}x = \dfrac{9}{{10}}\\
Do:{\sin ^2}x + {\cos ^2}x = 1\\
\Rightarrow {\sin ^2}x = \dfrac{1}{{10}}\\
\Rightarrow A = \dfrac{{3 - 4{{\sin }^2}x}}{{108{{\sin }^2}x - 9}} - 1\\
= \dfrac{{3 - 4.\dfrac{1}{{10}}}}{{108.\dfrac{1}{{10}} - 9}} - 1 = \dfrac{4}{9}
\end{array}$
