Đáp án: $x = \pm \dfrac{\pi }{4} + k\pi $
Giải thích các bước giải:
$\begin{array}{l}
\dfrac{{1 + {{\sin }^2}x}}{{1 - {{\sin }^2}x}} - 1\\
= \dfrac{{1 + {{\sin }^2}x - 1 + {{\sin }^2}x}}{{1 - {{\sin }^2}x}}\\
= \dfrac{{2{{\sin }^2}x}}{{{{\cos }^2}x}}\left( {do:co{s^2}x + {{\sin }^2}x = 1} \right)\\
= 2.ta{n^2}x\\
\Rightarrow \dfrac{{1 + {{\sin }^2}x}}{{1 - {{\sin }^2}x}} = 2{\tan ^2}x + 1\\
Do:\dfrac{{1 + {{\sin }^2}x}}{{1 - {{\sin }^2}x}} = 1 + 2{\cot ^2}x\\
\Rightarrow 2{\tan ^2}x + 1 = 2{\cot ^2}x + 1\\
\Rightarrow {\tan ^2}x = {\cot ^2}x\\
\Rightarrow \tan x = \cot x = \dfrac{1}{{{\mathop{\rm tanx}\nolimits} }}\\
\Rightarrow {\tan ^2}x = 1\\
\Rightarrow \left[ \begin{array}{l}
\tan x = 1\\
\tan x = - 1
\end{array} \right.\\
\Rightarrow x = \pm \dfrac{\pi }{4} + k\pi
\end{array}$
