Đáp án:
$\begin{array}{l}
a)\cot a = \dfrac{6}{7}\\
\Leftrightarrow \tan a = \dfrac{1}{{\cot a}} = \dfrac{7}{6}\\
\left\{ \begin{array}{l}
\dfrac{1}{{{{\cos }^2}a}} = {\tan ^2}a + 1\\
\dfrac{1}{{{{\sin }^2}a}} = {\cot ^2}a + 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{\cos ^2}a = \dfrac{{36}}{{85}}\\
{\sin ^2}a = \dfrac{{49}}{{85}}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\cos a = \pm \dfrac{{6\sqrt {85} }}{{85}}\\
\sin a = \pm \dfrac{{7\sqrt {85} }}{{85}}
\end{array} \right.\\
Vay\,\tan a = \dfrac{7}{6};\sin a = \pm \dfrac{{7\sqrt {85} }}{{85}};\cos a = \pm \dfrac{{6\sqrt {85} }}{{85}}\\
b)\sin a = \dfrac{8}{{17}}\\
Do:{\sin ^2}a + {\cos ^2}a = 1\\
\Leftrightarrow {\cos ^2}a = 1 - {\left( {\dfrac{8}{{17}}} \right)^2} = \dfrac{{225}}{{289}}\\
B = {\sin ^2}a + 3{\cos ^2}a\\
= {\sin ^2}a + {\cos ^2}a + 2{\cos ^2}a\\
= 1 + 2.\dfrac{{225}}{{289}}\\
= \dfrac{{739}}{{289}}
\end{array}$
