Bài 6.20 trang 186 SBT đại số 10

LG a

\(A = \tan {18^0}\tan {288^0} + \sin {32^0}\sin {148^0} \) \( - \sin {302^0}\sin {122^0}\)

Lời giải chi tiết:

\(A = \tan ({90^0} - {72^0})\tan ({360^0} - {72^0}) \) \(+ \sin {32^0}\sin ({180^0} - {32^0}) \) \( - \sin ({360^0} - {58^0})\sin ({180^0} - {58^0})\)

\(\eqalign{
& =\cot {72^0}( - \tan {72^0}) + {\sin ^2}{32^0} + {\sin ^2}{58^0} \cr 
& = - 1 + {\sin ^2}{32^0} + c{\rm{o}}{{\rm{s}}^2}{32^0} \cr 
& = - 1 + 1 = 0 \cr} \)

LG b

\(B = {{1 + {{\sin }^4}\alpha  - c{\rm{o}}{{\rm{s}}^4}\alpha } \over {1 - {{\sin }^6}\alpha  - c{\rm{o}}{{\rm{s}}^6}\alpha }}\)

Lời giải chi tiết:

\(\eqalign{
& B = {{1 + ({{\sin }^2}\alpha + c{\rm{o}}{{\rm{s}}^2}\alpha )(si{n^2}\alpha - c{\rm{o}}{{\rm{s}}^2}\alpha )} \over {1 - ({{\sin }^2}\alpha + c{\rm{o}}{{\rm{s}}^2}\alpha )({{\sin }^4}\alpha - {{\sin }^2}\alpha c{\rm{o}}{{\rm{s}}^2}\alpha + c{\rm{o}}{{\rm{s}}^4}\alpha )}} \cr 
& = {{1 + {{\sin }^2}\alpha - c{\rm{o}}{{\rm{s}}^2}\alpha } \over {1 - {\rm{[}}{{({{\sin }^2}\alpha + c{\rm{o}}{{\rm{s}}^2}\alpha )}^2} - 3{{\sin }^2}\alpha c{\rm{o}}{{\rm{s}}^2}\alpha }} \cr 
& = {{2{{\sin }^2}\alpha } \over {3{{\sin }^2}\alpha c{\rm{o}}{{\rm{s}}^2}\alpha }} = {2 \over 3}(1 + {\tan ^2}\alpha ) \cr} \)

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