Đáp án:
$\begin{array}{l}
a)\tan A = 1,5\\
\Leftrightarrow \cot A = \dfrac{1}{{\tan A}} = \dfrac{2}{3}\\
+ Do:\dfrac{1}{{{{\cos }^2}A}} = {\tan ^2}A + 1 = \dfrac{{13}}{4}\\
\Leftrightarrow {\cos ^2}A = \dfrac{4}{{13}}\\
\Leftrightarrow \cos A = \pm \dfrac{{2\sqrt {13} }}{{13}}\\
+ \sin A = \cos A.\tan A = \pm \dfrac{{3\sqrt {13} }}{{13}}\\
Vậy\,\cot A = \dfrac{2}{3};\cos A = \pm \dfrac{{2\sqrt {13} }}{{13}};\sin A = \pm \dfrac{{3\sqrt {13} }}{{13}}\\
b)\cos A = 0,6\\
Do:{\sin ^2}A + {\cos ^2}A = 1\\
\Leftrightarrow {\sin ^2}A = 0,64\\
\Leftrightarrow \sin A = \pm 0,8\\
\Leftrightarrow \left\{ \begin{array}{l}
\tan A = \dfrac{{\sin A}}{{\cos A}} = \pm \dfrac{4}{3}\\
\cot A = \pm \dfrac{3}{4}
\end{array} \right.
\end{array}$
