Đáp án:

$D = R\backslash \left\{ { - {\pi \over 2} + k2\pi ,\,\,k \in Z} \right\} $Lời giải:

\(y = \sqrt {{{1 - \sin x} \over {1 + \sin x}}} \)

Hàm số xác định khi

\(\eqalign{ & y = \sqrt {{{1 - \sin x} \over {1 + \sin x}}} \cr & \Leftrightarrow \left\{ \matrix{ {{1 - \sin x} \over {1 + \sin x}} \ge 0 \hfill \cr 1 + \sin x \ne 0\,\,\left( * \right) \hfill \cr} \right. \cr & \text{Ta có: }- 1 \le \sin x \le 1 \cr & \Leftrightarrow \left\{ \matrix{ 1 - \sin x \ge 0 \hfill \cr 1 + \sin x \ge 0 \hfill \cr} \right. \cr & \Rightarrow {{1 - \sin x} \over {1 + \sin x}} \ge 0\text{ luôn đúng }\forall x\text{ thảo mãn }\left( * \right) \cr & \left( * \right) \Leftrightarrow \sin x \ne - 1 \Leftrightarrow x \ne - {\pi \over 2} + k2\pi \,\,\left( {k \in Z} \right) \cr & \Rightarrow D = R\backslash \left\{ { - {\pi \over 2} + k2\pi ,\,\,k \in Z} \right\} \cr} \).