Cho $\sin \alpha =\frac{1}{3}$ với $90°<\alpha <180°$.

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 $90°<\alpha <180°$ 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 \[\frac{{6\sin \alpha  + 3\sqrt 2 \cos \alpha }}{{2\sqrt 2 \tan \alpha  + \sqrt 2 \cot \alpha }} = \frac{{6 \cdot \frac{1}{3} + 3\sqrt 2  \cdot \left( { - \frac{{2\sqrt 2 }}{3}} \right)}}{{2\sqrt 2  \cdot \left( { - \frac{{\sqrt 2 }}{4}} \right) + \sqrt 2  \cdot \left( { - 2\sqrt 2 } \right)}} = \frac{2}{5}\].