Đáp án:
$\begin{array}{l}
y = \sqrt {\dfrac{{\sin x - 1}}{{\cos x + 1}}} \\
Dkxd:\dfrac{{\sin x - 1}}{{\cos x + 1}} \ge 0\\
\Leftrightarrow \left\{ \begin{array}{l}
\sin x - 1 \ge 0\\
\cos x\# - 1\left( {do:\cos x \ge - 1} \right)
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\sin x \ge 1\\
x\# \pi + k2\pi
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\sin x = 1\left( {do:\sin x \le 1} \right)\\
x\# \pi + k2\pi
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{\pi }{2} + k2\pi \\
x\# \pi + k2\pi
\end{array} \right.\\
\Leftrightarrow x = \dfrac{\pi }{2} + k2\pi \\
Vay\,x = \dfrac{\pi }{2} + k2\pi \\
y = \dfrac{1}{{\left( {2{{\cos }^2}x - 1} \right).\tan x}}\\
Dkxd:\left\{ \begin{array}{l}
2{\cos ^2}x - 1\# 0\\
\tan x\# 0\\
\cos x\# 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{\cos ^2}x\# \dfrac{1}{2}\\
\sin x\# 0\\
\cos x\# 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x\# \pm \dfrac{\pi }{4} + k2\pi \\
x\# \dfrac{{k\pi }}{2}
\end{array} \right.\\
Vay\,x\# \dfrac{{k\pi }}{2};x\# \pm \dfrac{\pi }{4} + k2\pi
\end{array}$
