$\left\{ \begin{array}{l}1 - {\tan ^2}x \ne 0\\\cos 2x \ne 0\\\cos x \ne 0\end{array} \right. $ $\Leftrightarrow \left\{ \begin{array}{l}\dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\cos }^2}x}} \ne 0\\\cos 2x \ne 0\\\cos x \ne 0\end{array} \right. $ $\Leftrightarrow \left\{ \begin{array}{l}\cos 2x \ne 0\\\cos x \ne 0\end{array} \right. $ $\Leftrightarrow \left\{ \begin{array}{l}2x \ne \dfrac{\pi }{2} + k\pi \\x \ne \dfrac{\pi }{2} + k\pi \end{array} \right. $ $\Leftrightarrow \left\{ \begin{array}{l}x \ne \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\x \ne \dfrac{\pi }{2} + k\pi \end{array} \right.\,\,\,\left( {k \in Z} \right)$
$\begin{array}{l}\dfrac{{{a^2}}}{{1 - {{\tan }^2}x}} = \dfrac{{{{\sin }^2}x + {a^2} - 2}}{{\cos 2x}} \Leftrightarrow \dfrac{{{a^2}}}{{\dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\cos }^2}x}}}} = \dfrac{{{{\sin }^2}x + {a^2} - 2}}{{\cos 2x}}\\ \Leftrightarrow \dfrac{{{a^2}{{\cos }^2}x}}{{\cos 2x}} = \dfrac{{{{\sin }^2}x + {a^2} - 2}}{{\cos 2x}} \Leftrightarrow {a^2}{\cos ^2}x = {\sin ^2}x + {a^2} - 2\\ \Leftrightarrow {a^2}{\cos ^2}x = 1 - {\cos ^2}x + {a^2} - 2\\ \Leftrightarrow \left( {{a^2} + 1} \right){\cos ^2}x = {a^2} - 1 \Leftrightarrow {\cos ^2}x = \dfrac{{{a^2} - 1}}{{{a^2} + 1}} < 1\end{array}$
Vì \(\cos x \ne 0 \Rightarrow 0 < {\cos ^2}x \le 1 \Leftrightarrow {\cos ^2}x > 0 \Leftrightarrow {a^2} - 1 > 0 \Rightarrow \left| a \right| > 1\)
