Đáp án:
$\begin{array}{l}
B1)0 < a < {90^0}\\
\Leftrightarrow 0 < \sin a,\cos a < 1\\
a)\sin a = \dfrac{1}{3}\\
\Leftrightarrow \cos a = \sqrt {1 - {{\sin }^2}a} = \dfrac{{2\sqrt 2 }}{3}\\
\Leftrightarrow \tan a = \dfrac{{\sin a}}{{\cos a}} = \dfrac{{\dfrac{1}{3}}}{{\dfrac{{2\sqrt 2 }}{3}}} = \dfrac{{\sqrt 2 }}{4}\\
b)\cos a = \dfrac{2}{{\sqrt 5 }}\\
\Leftrightarrow \sin a = \sqrt {1 - {{\cos }^2}a} = \dfrac{1}{{\sqrt 5 }}\\
\Leftrightarrow \tan a = \dfrac{{\sin a}}{{\cos a}} = \dfrac{{\dfrac{1}{{\sqrt 5 }}}}{{\dfrac{2}{{\sqrt 5 }}}} = \dfrac{1}{2}\\
B2)\\
\cot a = \dfrac{1}{2}\\
\Leftrightarrow \dfrac{1}{{{{\sin }^2}a}} = {\cot ^2}a + 1 = \dfrac{5}{4}\\
\Leftrightarrow \sin a = \dfrac{2}{{\sqrt 5 }}\\
\Leftrightarrow \cos a = \sin a.\cot a = \dfrac{2}{{\sqrt 5 }}.\dfrac{1}{2} = \dfrac{1}{{\sqrt 5 }}\\
S = 3{\sin ^2}a + 4\sin a.\cos a + 5{\cos ^2}a\\
= 3.{\left( {\dfrac{2}{{\sqrt 5 }}} \right)^2} + 4.\dfrac{2}{{\sqrt 5 }}.\dfrac{1}{{\sqrt 5 }} + 5.{\left( {\dfrac{1}{{\sqrt 5 }}} \right)^2}\\
= \dfrac{{12}}{5} + \dfrac{8}{5} + 1\\
= 5
\end{array}$
