Đáp án + Giải thích các bước giải:

a) Ta có: $0 < \alpha < \dfrac{\pi}{2}$

$\Leftrightarrow \dfrac{\pi}{4} < \alpha + \dfrac{\pi}{4} < \dfrac{3\pi}{4}$

$\Leftrightarrow \begin {cases} \sin \bigg(\alpha + \dfrac{\pi}{4}\bigg) > 0, \hspace{1cm} 0 < \alpha < \dfrac{\pi}{4} \\ \sin \bigg(\alpha + \dfrac{\pi}{4}\bigg) < 0, \hspace{1cm} \dfrac{\pi}{4} < \alpha < \dfrac{\pi}{2} \end {cases}$

b) Ta có: $0 < \alpha < \dfrac{\pi}{2}$

$\Leftrightarrow -\dfrac{\pi}{2} < \alpha - \dfrac{\pi}{2} < 0$

$\Leftrightarrow \tan \bigg(\alpha - \dfrac{\pi}{2}\bigg)  < 0$