Biết \(\tan \alpha  = 2\) và \(0^\circ  < \alpha  < 90^\circ \).

a) \(\cot \alpha  = \frac{1}{2}\).

b) \[\sin \alpha  \cdot \cos \alpha  < 0\]. 

c) \[\cos \alpha  = \frac{1}{{\sqrt 5 }}\].

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

a) Đúng. Ta có \(\cot \alpha  = \frac{1}{{\tan \alpha }} = \frac{1}{2}\).

b) Sai. \(\tan \alpha  = \frac{{\sin \alpha }}{{{\rm{cos}}\alpha }} = 2 > 0 \Rightarrow \sin \alpha  \cdot {\rm{cos}}\alpha  > 0\).

c) Đúng. Vì \(0^\circ  < \alpha  < 90^\circ \) nên \({\rm{cos}}\alpha  > 0\).

Ta có \(1 + {\tan ^2}\alpha  = \frac{1}{{{\rm{co}}{{\rm{s}}^2}\alpha }} \Rightarrow {\rm{co}}{{\rm{s}}^2}\alpha  = \frac{1}{{1 + {2^2}}} = \frac{1}{5} \Rightarrow {\rm{cos}}\alpha  = \frac{{\sqrt 5 }}{5} = \frac{1}{{\sqrt 5 }}\).

d) Sai. Ta có \(\tan \alpha  = \frac{{\sin \alpha }}{{{\rm{cos}}\alpha }} \Rightarrow \sin \alpha  = \tan \alpha  \cdot {\rm{cos}}\alpha  = \frac{{2\sqrt 5 }}{5}\).

Suy ra \({\rm{sin}}\alpha \,{\rm{ + }}\,{\rm{cos}}\alpha  = \frac{{2\sqrt 5 }}{5} + \frac{{\sqrt 5 }}{5} = \frac{{3\sqrt 5 }}{5}\).