Gợi ý làm bài
a) \({\sin ^2}({180^0} - \alpha ) + ta{n^2}({180^0} - \alpha ){\tan ^2}({270^0} - \alpha ) + \sin ({90^0} + \alpha )cos(\alpha - {360^0})\)
= \({\sin ^2}\alpha + {\tan ^2}\alpha {\cot ^2}\alpha + {\cos ^2}\alpha = 2\)
b) \({{\cos (\alpha - {{90}^0})} \over {\sin ({{180}^0} - \alpha )}} + {{\tan (\alpha - {{180}^0})c{\rm{os(18}}{{\rm{0}}^0} + \alpha )\sin ({{270}^0} + \alpha )} \over {\tan ({{270}^0} + \alpha )}}\)
= \({{\sin \alpha } \over {\cos \alpha }} + {{\tan \alpha ( - \cos \alpha )( - \cos \alpha )} \over { - \cot \alpha }} = 1 - {\sin ^2}\alpha = {\cos ^2}\alpha \)
c) \({{\cos ( - {{288}^0})cot{{72}^0}} \over {tan( - {{162}^0})\sin {{108}^0}}} + \tan {18^0}\)
\( = {{\cos ({{72}^0} - {{360}^0})\cot {{72}^0}} \over {\tan ({{18}^0} - {{180}^0})\sin ({{180}^0} - {{72}^0})}} - \tan {18^0}\)
= \({{{\rm{cos7}}{{\rm{2}}^0}\cot {{72}^0}} \over {\tan {{18}^0}\sin {{72}^0}}} - \tan {18^0}\)
= \({{{{\cot }^2}{{72}^0}} \over {\tan {{18}^0}}} - \tan {18^0} = {{{{\tan }^2}{{18}^0}} \over {\tan {{18}^0}}} - \tan {18^0} = 0\)
d) Ta có: \(\sin {70^0} = \cos {20^0},\sin {50^0} = cos4{{\rm{0}}^0};\sin {40^0} = cos{50^0}\). Vì vậy
\({{\sin {{20}^0}\sin {\rm{3}}{{\rm{0}}^0}\sin {{40}^0}\sin {{50}^0}\sin {{60}^0}\sin {{70}^0}} \over {cos{{10}^0}{\rm{cos5}}{{\rm{0}}^0}}}\)
= \(\eqalign{
& {{{1 \over 2}.{{\sqrt 3 } \over 2}.\sin {{20}^0}\cos {\rm{2}}{{\rm{0}}^0}\cos {{50}^0}\cos {{40}^0}} \over {cos{{10}^0}{\rm{cos5}}{{\rm{0}}^0}}} \cr
& = {{{1 \over 2}.{{\sqrt 3 } \over 4}\sin {{40}^0}.cos{{40}^0}} \over {{\rm{cos1}}{{\rm{0}}^0}}} \cr} \)
= \({{{{\sqrt 3 } \over {16}}\sin {{80}^0}} \over {cos{{10}^0}}} = {{\sqrt 3 } \over {16}}\)
