cot α = − 1/2 .

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.