Bài tập 6.50 trang 192 SBT Toán 10

a) \({\sin ^2}\left( {{{180}^0} - \alpha } \right) + {\tan ^2}\left( {{{180}^0} - \alpha } \right){\tan ^2}\left( {{{270}^0} + \alpha } \right) + \sin \left( {{{90}^0} + \alpha } \right)\cos \left( {\alpha  - {{360}^0}} \right)\)

\( = {\sin ^2}\alpha  + {\tan ^2}\alpha {\cot ^2}\alpha  + {\cos ^2}\alpha  = 2\)

b) \(\frac{{\cos \left( {\alpha  - {{90}^0}} \right)}}{{\sin \left( {{{180}^0} - \alpha } \right)}} + \frac{{\tan \left( {\alpha  - {{180}^0}} \right)\cos \left( {{{180}^0} + \alpha } \right)\sin \left( {{{270}^0} + \alpha } \right)}}{{\tan \left( {{{270}^0} + \alpha } \right)}}\)

\( = \frac{{\sin \alpha }}{{\sin \alpha }} + \frac{{\tan \alpha \left( { - \cos \alpha } \right)\left( { - \cos \alpha } \right)}}{{ - \cot \alpha }} = 1 - {\sin ^2}\alpha  = {\cos ^2}\alpha \)

c) \(\frac{{\cos \left( { - {{288}^0}} \right)\cot {{72}^0}}}{{\tan \left( { - {{162}^0}} \right)\sin {{108}^0}}} - \tan {18^0}\)

\(\begin{array}{l}
 = \frac{{\cos \left( {{{72}^0} - {{360}^0}} \right)\cot {{72}^0}}}{{\tan \left( {{{18}^0} - {{180}^0}} \right)\sin \left( {{{180}^0} - {{72}^0}} \right)}} - \tan {18^0}\\
 = \frac{{\cos {{72}^0}\cot {{72}^0}}}{{\tan {{18}^0}\sin {{72}^0}}} - \tan {18^0}\\
 = \frac{{{{\cot }^2}{{72}^0}}}{{\tan {{18}^0}}} - \tan {18^0} = \frac{{{{\tan }^2}{{18}^0}}}{{\tan {{18}^0}}} - \tan {18^0} = 0
\end{array}\)

d) Ta có \(\sin {70^0} = \cos {20^0},\sin {50^0} = \cos {40^0},\sin {40^0} = \cos {50^0}\). Vì vậy 

\(\frac{{\sin {{20}^0}\sin {{30}^0}\sin {{40}^0}\sin {{50}^0}\sin {{60}^0}\sin {{70}^0}}}{{\cos {{10}^0}\cos {{50}^0}}}\)

\(\begin{array}{l}
\frac{{\frac{1}{2}.\frac{{\sqrt 3 }}{2}.\sin {{20}^0}\cos {{20}^0}\cos {{50}^0}\cos {{40}^0}}}{{\cos {{10}^0}\cos {{50}^0}}} = \frac{{\frac{1}{2}.\frac{{\sqrt 3 }}{4}.\sin {{40}^0}\cos {{40}^0}}}{{\cos {{10}^0}}}\\
 = \frac{{\frac{{\sqrt 3 }}{{16}}\sin {{80}^0}}}{{\cos {{10}^0}}} = \frac{{\sqrt 3 }}{6}
\end{array}\)

-- Mod Toán 10 HỌC247