Đáp án: $\left[ \begin{array}{l}
B = \dfrac{{2\sqrt 2 - 1}}{3}\\
B = \dfrac{{ - 2\sqrt 2 - 1}}{3}
\end{array} \right.$
Giải thích các bước giải:
$\begin{array}{l}
Do:\dfrac{1}{{{{\cos }^2}a}} = {\tan ^2}a + 1 = 3\\
\Leftrightarrow {\cos ^2}a = \dfrac{1}{3}\\
\Leftrightarrow \cos a = \pm \dfrac{{\sqrt 3 }}{3}\\
Do:{\sin ^2}a + {\cos ^2}a = 1\\
\Leftrightarrow {\sin ^2}a = 1 - {\cos ^2}a = 1 - \dfrac{1}{3} = \dfrac{2}{3}\\
\Leftrightarrow \sin a = \pm \dfrac{{\sqrt 6 }}{3}\\
B = {\sin ^2}a + 2\sin a.\cos a - 3{\cos ^2}a\\
= \left[ \begin{array}{l}
\dfrac{2}{3} - 3.\dfrac{1}{3} + 2.\dfrac{{\sqrt 6 }}{3}.\dfrac{{\sqrt 3 }}{3} = \dfrac{{2\sqrt 2 - 1}}{3}\\
\dfrac{2}{3} - 3.\dfrac{1}{3} - 2.\dfrac{{\sqrt 6 }}{3}.\dfrac{{\sqrt 3 }}{3} = \dfrac{{ - 2\sqrt 2 - 1}}{3}
\end{array} \right.
\end{array}$
