2 câu trả lời
$\tan^2x+1\neq 0\Leftrightarrow \tan^2x\neq -1$ (luôn đúng)
Mà $\tan^2x=\dfrac{\sin^2x}{\cos^2x}$
$\Rightarrow $ ĐK: $\cos x\neq 0$
$\Leftrightarrow x\neq \dfrac{\pi}{2}+k\pi$
$D=\mathbb{R}$ \ $\{\dfrac{\pi}{2}+k\pi\}$
$$\eqalign{ & y = {{\cos x} \over {{{\tan }^2}x + 1}} \cr & Do\,\,{\tan ^2}x \ge 0 \Leftrightarrow {\tan ^2}x + 1 > 0 \cr & \Rightarrow DKXD:\,\,\cos x \ne 0 \Leftrightarrow x \ne {\pi \over 2} + k\pi \cr & \Rightarrow D = R\backslash \left\{ {{\pi \over 2} + k\pi ,\,\,k \in Z} \right\} \cr} $$