Bài 34 trang 207 SGK Đại số 10 Nâng cao

Đề bài

Chứng minh rằng:

a) \({{1 - 2\sin \alpha \,\cos \alpha } \over {{{\cos }^2}\alpha  - {{\sin }^2}\alpha }} = {{1 - \tan \alpha } \over {1 + \tan \alpha }}\) khi các biểu thức đó có nghĩa

b) \(ta{n^2}\alpha {\rm{ }} - {\rm{ }}si{n^2}\alpha {\rm{ }} = {\rm{ }}ta{n^2}\alpha {\rm{ }}si{n^2}\alpha \)

c) \(2(1{\rm{ }}-\sin\alpha {\rm{ }})\left( {1{\rm{ }} + {\rm{ }}cos\alpha } \right){\rm{ }} = {\rm{ }}{\left( {1{\rm{ }} - {\rm{ }}\sin\alpha {\rm{ }} + {\rm{ }}\cos\alpha {\rm{ }}} \right)^2}\)

Hướng dẫn giải

a) Ta có:

\(\eqalign{
& {{1 - 2\sin \alpha \,\cos \alpha } \over {{{\cos }^2}\alpha - {{\sin }^2}\alpha }} = {{{{(cos\alpha - \sin \alpha )}^2}} \over {(cos\alpha - \sin \alpha )(cos\alpha + \sin \alpha )}} \cr
& = {{(cos\alpha - \sin \alpha )} \over {(cos\alpha + \sin \alpha )}} = {{\cos \alpha (1 - \tan \alpha )} \over {\cos \alpha (1 + tan\alpha )}} \cr
& = {{1 - \tan \alpha } \over {1 + \tan \alpha }} \cr} \) 

b) Ta có:

\(ta{n^2}\alpha {\rm{ }} - {\rm{ }}si{n^2}\alpha {\rm{ }} = {\rm{ }}ta{n^2}\alpha ({\rm{ }}1 - {\rm{ }}co{s^2}\alpha ){\rm{ }} = {\rm{ }}ta{n^2}\alpha {\rm{ }}si{n^2}\alpha \)

c) Ta có:

\(2(1-si{n}\alpha {\rm{ }})\left( {1{\rm{ }} + {\rm{ }}cos\alpha } \right){\rm{ }}\)

\(= {\rm{ }}2{\rm{ }}-{\rm{ }}2sin\alpha {\rm{ }} + {\rm{ }}2cos\alpha {\rm{ }}-{\rm{ }}2sin\alpha {\rm{ }}cos\alpha \)

\( = {\rm{ }}1{\rm{ }} + {\rm{ }}si{n^2}\alpha {\rm{ }} + {\rm{ }}co{s^2}\alpha {\rm{ }}-{\rm{ }}2sin\alpha {\rm{ }} + {\rm{ }}2cos\alpha {\rm{ }} \)

    \(- {\rm{ }}2sin\alpha {\rm{ }}cos\alpha \)

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