Bài 34 trang 207 Đại số 10 Nâng cao: Chứng minh rằng:...

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

Đáp án

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:

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

$= {\rm{ }}2{\rm{ }}-{\rm{ }}2sin\alpha {\rm{ }} + {\rm{ }}2cos\alpha {\rm{ }}-{\rm{ }}2sin\alpha {\rm{ }}cos\alpha$

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

    $- {\rm{ }}2sin\alpha {\rm{ }}cos\alpha$

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