Tìm txd Y=√1-sinx Tan^2x

\(y=\sqrt{1-\sin x}{\tan}^2x\) ĐKXĐ: \(\left\{ \begin{array}{l} 1-\sin x\ge 0 \\ \cos x\ne0 \end{array} \right .\) \(\Leftrightarrow \left\{ \begin{array}{l} \sin x\le 1\text{ luôn đúng} \\ x\ne\dfrac{\pi}{2}+k\pi,(k\mathbb Z) \end{array} \right .\) suy ra ĐKXĐ là \(x\ne\dfrac{\pi}{2}+k\pi,(k\in\mathbb Z)\). \(y=\dfrac{\sqrt{1-\sin x}}{{\tan}^2x}\) ĐKXĐ: \(\left\{ \begin{array}{l} 1-\sin x\ge 0 \\\sin x\ne0\\ \cos x\ne0 \end{array} \right .\) \(\Leftrightarrow \left\{ \begin{array}{l} \sin x\le 1\text{ luôn đúng} \\ \sin 2x\ne0 \end{array} \right .\) \(\Leftrightarrow 2x\ne k\pi\) \(\Leftrightarrow x\ne k\dfrac{\pi}{2},(k\in\mathbb Z)\) suy ra ĐKXĐ là \(x\ne k\dfrac{\pi}{2},(k\in\mathbb Z)\).