Đáp án
a)
\(\eqalign{
& {{\sin \alpha - \sin \beta } \over {\cos \alpha - \cos \beta }} = {{2\cos {{\alpha + \beta } \over 2}\sin {{\alpha - \beta } \over 2}} \over { - 2\sin {{\alpha + \beta } \over 2}\sin {{\alpha - \beta } \over 2}}} \cr
& = - \cot {{\alpha + \beta } \over 2} = - \cot {\pi \over 6} = - \sqrt 3 \cr} \)
b)
\({{\cos \alpha - \cos 7\alpha } \over {\sin 7\alpha - sin\alpha }} = {{2\sin 4\alpha \sin 3\alpha } \over {2\cos 4\alpha \sin 3\alpha }} = \tan 4\alpha \)