Bài tập 6.20 trang 186 SBT Toán 10

\(\begin{array}{l}
A = \tan \left( {{{90}^0} - {{72}^0}} \right)\tan \left( {{{360}^0} - {{72}^0}} \right) + \sin {32^0}\sin \left( {{{180}^0} - {{32}^0}} \right) - \sin \left( {{{360}^0} - {{58}^2}} \right)\sin \left( {{{180}^0} - {{58}^0}} \right)\\
 = \cot {72^2}\left( { - \tan {{72}^0}} \right) + {\sin ^2}{32^0} + {\sin ^2}{58^0}\\
 =  - 1 + {\sin ^2}{32^0} + {\cos ^2}{32^0} =  - 1 + 1 = 0
\end{array}\)

\(\begin{array}{l}
B = \frac{{1 + {{\sin }^4}\alpha  - {{\cos }^4}\alpha }}{{1 - {{\sin }^6}\alpha  - {{\cos }^6}\alpha }} = \frac{{1 + \left( {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right)\left( {{{\sin }^2}\alpha  - {{\cos }^2}\alpha } \right)}}{{1 - \left( {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right)\left( {{{\sin }^4}\alpha  - {{\sin }^2}\alpha {{\cos }^2}\alpha  + {{\cos }^4}\alpha } \right)}}\\
 = \frac{{1 + {{\sin }^2}\alpha  - {{\cos }^2}\alpha }}{{1 - \left[ {{{\left( {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right)}^2} - 3{{\sin }^2}\alpha {{\cos }^2}\alpha } \right]}}\\
 = \frac{{3{{\sin }^2}\alpha }}{{3{{\sin }^2}\alpha {{\cos }^2}\alpha }} = \frac{2}{3}\left( {1 + {{\tan }^2}\alpha } \right)
\end{array}\)

-- Mod Toán 10 HỌC247