Cho\(\sin \alpha = \frac{1}{3}\) với \(90^\circ < \alpha < 180^\circ \).

a) Giá trị \(\sin \alpha \cdot \cos \alpha < 0\).

b) \(\cos \alpha = - \frac{{2\sqrt 2 }}{3}.\)

c) \(\tan \alpha = \frac{{\sqrt 2 }}{4}.\)

d) \[\frac{{6\sin \alpha + 3\sqrt 2 \cos \alpha }}{{2\sqrt 2 \tan \alpha + \sqrt 2 \cot \alpha }} = \frac{2}{5}.\]

a) Đúng. Ta có \(\sin \alpha = \frac{1}{3} > 0\).

Do \(90^\circ < \alpha < 180^\circ \) nên \(\cos \alpha < 0\). Vậy giá trị \(\sin \alpha \cdot \cos \alpha < 0\).

b) Đúng. Vì \(\cos \alpha < 0\), mà \({\sin ^2}\alpha + {\cos ^2}\alpha = 1\), suy ra \(\cos \alpha = - \sqrt {1 - {{\sin }^2}\alpha } = - \sqrt {1 - \frac{1}{9}} = - \frac{{2\sqrt 2 }}{3}\).

c) Sai. Ta có \(\tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }} = \frac{{\frac{1}{3}}}{{ - \frac{{2\sqrt 2 }}{3}}} = - \frac{1}{{2\sqrt 2 }} = - \frac{{\sqrt 2 }}{4}\).

d) Đúng. Ta có \(\cot \alpha = \frac{1}{{\tan \alpha }} = \frac{1}{{ - \frac{{\sqrt 2 }}{4}}} = - 2\sqrt 2 .\)

Vậy \[\frac{{6\sin \alpha + 3\sqrt 2 \cos \alpha }}{{2\sqrt 2 \tan \alpha + \sqrt 2 \cot \alpha }} = \frac{{6 \cdot \frac{1}{3} + 3\sqrt 2 \cdot \left( { - \frac{{2\sqrt 2 }}{3}} \right)}}{{2\sqrt 2 \cdot \left( { - \frac{{\sqrt 2 }}{4}} \right) + \sqrt 2 \cdot \left( { - 2\sqrt 2 } \right)}} = \frac{2}{5}\].