Đáp án:
\(\left[ \begin{array}{l}
N = \dfrac{{\sqrt {13} }}{3}\\
N = - \dfrac{{\sqrt {13} }}{3}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
\tan a = \dfrac{2}{3} \to \cot a = \dfrac{3}{2}\\
\to \sin a = \dfrac{2}{3}\cos a\\
{\sin ^2}a + {\cos ^2}a = 1\\
\to \dfrac{4}{9}{\cos ^2}a + {\cos ^2}a = 1\\
\to \dfrac{{13}}{9}{\cos ^2}a = 1\\
\to {\cos ^2}a = \dfrac{9}{{13}}\\
\to \left[ \begin{array}{l}
\cos a = \dfrac{3}{{\sqrt {13} }}\\
\cos a = - \dfrac{3}{{\sqrt {13} }}
\end{array} \right. \to \left[ \begin{array}{l}
\sin a = \dfrac{2}{{\sqrt {13} }}\\
\sin a = - \dfrac{2}{{\sqrt {13} }}
\end{array} \right.\\
\left[ \begin{array}{l}
N = \dfrac{1}{{\sin a.\cot a}} = \dfrac{1}{{\dfrac{2}{{\sqrt {13} }}.\dfrac{3}{2}}}\\
N = \dfrac{1}{{\sin a.\cot a}} = - \dfrac{1}{{\dfrac{2}{{\sqrt {13} }}.\dfrac{3}{2}}}
\end{array} \right.\\
\to \left[ \begin{array}{l}
N = \dfrac{{\sqrt {13} }}{3}\\
N = - \dfrac{{\sqrt {13} }}{3}
\end{array} \right.
\end{array}\)
