Gợi ý làm bài
a)
\(\eqalign{
& \sin 6\alpha \cot 3\alpha - c{\rm{os6}}\alpha \cr
& = 2\sin 3\alpha \cos 3\alpha .{{\cos 3\alpha } \over {\sin 3\alpha }} - (2{\cos ^2}3\alpha - 1) \cr} \)
= \(2{\cos ^2}3\alpha - 2{\cos ^2}3\alpha + 1 = 1\)
b)
\({{\rm{[}}\tan ({90^0} - \alpha ) - \cot ({90^0} + \alpha ){\rm{]}}^2} - {{\rm{[}}c{\rm{ot(18}}{{\rm{0}}^0} + \alpha ) + \cot ({270^0} + \alpha ){\rm{]}}^2}\)
= \({(\cot \alpha + \tan \alpha )^2} - {(\cot \alpha - \tan \alpha )^2}\)
= \({\cot ^2}\alpha + 2 + {\tan ^2}\alpha - {\cot ^2}\alpha + 2 - {\tan ^2}\alpha = 4\)
c)
\(\eqalign{
& (\tan \alpha - \tan \beta )cot(\alpha - \beta ) - \tan \alpha \tan \beta \cr
& = {{\tan \alpha - \tan \beta } \over {\tan (\alpha - \beta )}} - \tan \alpha \tan \beta \cr} \)
=\(1 + \tan \alpha \tan \beta - \tan \alpha \tan \beta = 1\)
d)
\(\eqalign{
& (\cot {\alpha \over 3} - \tan {\alpha \over 3})\tan {{2\alpha } \over 3} \cr
& = ({{\cos {\alpha \over 3}} \over {\sin {\alpha \over 3}}} - {{\sin {\alpha \over 3}} \over {\cos {\alpha \over 3}}}){{\sin {{2\alpha } \over 3}} \over {\cos {{2\alpha } \over 3}}} \cr} \)
= \(\eqalign{
& {{{{\cos }^2}{\alpha \over 3} - {{\sin }^2}{\alpha \over 3}} \over {\sin {\alpha \over 3}\cos {\alpha \over 3}}}.{{\sin {{2\alpha } \over 3}} \over {\cos {{2\alpha } \over 3}}} \cr
& = {{\cos {{2\alpha } \over 3}} \over {{1 \over 2}\sin {{2\alpha } \over 3}}}.{{\sin {{2\alpha } \over 3}} \over {\cos {{2\alpha } \over 3}}} = 2 \cr} \)
