Biết tan α = 2 .

Ta có \(\cot \alpha  = \frac{1}{{\tan \alpha }} = \frac{1}{2}\); \(\cos 2\alpha  = 2{\cos ^2}\alpha  - 1 = \frac{2}{{{{\tan }^2}\alpha  + 1}} - 1 = \frac{2}{5} - 1 =  - \frac{3}{5}\);

\(\sin 2\alpha  = 2\sin \alpha \cos \alpha  = 2\tan \alpha {\cos ^2}\alpha  = \frac{{2\tan \alpha }}{{{{\tan }^2}\alpha  + 1}} = \frac{4}{5};\tan 2\alpha  = \frac{{\sin 2\alpha }}{{\cos 2\alpha }} =  - \frac{4}{3}\).

Đáp án:       a) Sai,                    b) Đúng,      c) Đúng,      d) Đúng.