Bài 6.12 trang 189 SBT đại số 10

Chứng minh các đẳng thức

a) \({{\tan \alpha  - \tan \beta } \over {{\rm{cot}}\beta {\rm{ - cot}}\alpha }} = \tan \alpha \tan \beta\)

b) \(\tan {100^0} + {{\sin {{530}^0}} \over {1 + \sin {{640}^0}}} = {1 \over {\sin {{10}^0}}}\)

c) \(2({\sin ^6}\alpha  + c{\rm{o}}{{\rm{s}}^6}\alpha ) + 1 = 3({\sin ^4}\alpha  + c{\rm{o}}{{\rm{s}}^4}\alpha )\)

Gợi ý làm bài

a)\(\eqalign{

& {{\tan \alpha - \tan \beta } \over {{\rm{cot}}\beta {\rm{ - cot}}\alpha }} = {{\tan \alpha - \tan \beta } \over {{1 \over {\tan \beta }} - {1 \over {\tan \alpha }}}} \cr
& = {{\tan \alpha - \tan \beta } \over {{{\tan \alpha - \tan \beta } \over {tan\alpha \tan \beta }}}} = \tan \alpha \tan \beta \cr} \) 

b)

\(\eqalign{
& \tan {100^0} + {{\sin {{530}^0}} \over {1 + \sin {{640}^0}}} \cr
& = \tan ({90^0} + {10^0}) + {{\sin ({{360}^0} + {{170}^0})} \over {1 + \sin ({{720}^0} - {{80}^0})}} \cr} \)

\(\eqalign{
& = - \cot {10^0} + {{\sin {{170}^0}} \over {1 - \sin {{80}^0}}} \cr
& = - {{\cos {{10}^0}} \over {\sin {{10}^0}}} + {{\sin {{10}^0}} \over {1 - c{\rm{os1}}{{\rm{0}}^0}}} \cr} \)

\( = {{ - \cos {{10}^0} + {{\cos }^2}{{10}^0} + {{\sin }^2}{{10}^0}} \over {\sin {{10}^0}(1 - c{\rm{os1}}{{\rm{0}}^0})}} = {1 \over {\sin {{10}^0}}}\)

\(\eqalign{
& c)2({\sin ^6}\alpha + c{\rm{o}}{{\rm{s}}^6}\alpha ) + 1 \cr
& = 2({\sin ^2}x + c{\rm{o}}{{\rm{s}}^2}x)({\sin ^4}x - {\sin ^2}x{\cos ^2}x + c{\rm{o}}{{\rm{s}}^4}x) + 1 \cr
& = 2({\sin ^4}x + c{\rm{o}}{{\rm{s}}^4}x) + {({\sin ^2}x + c{\rm{o}}{{\rm{s}}^2}x)^2} - 2{\sin ^{^2}}x{\cos ^2}x \cr
& = 2({\sin ^4}x + c{\rm{o}}{{\rm{s}}^4}x) + ({\sin ^4}x + c{\rm{o}}{{\rm{s}}^4}x) \cr
& = 3({\sin ^4}\alpha + c{\rm{o}}{{\rm{s}}^4}\alpha ) \cr} \)