Bài 6.34 trang 196 SBT đại số 10

Chứng minh các đẳng thức

a) \(\tan 3\alpha  - \tan 2\alpha  - \tan \alpha  = \tan \alpha \tan 2\alpha \tan 3\alpha \)

b) \({{4\tan \alpha (1 - {{\tan }^2}\alpha )} \over {{{(1 + {{\tan }^2}\alpha )}^2}}} = \sin 4\alpha \)

c) \({{1 + {{\tan }^4}\alpha } \over {{{\tan }^2}\alpha  + {{\cot }^2}\alpha }} = {\tan ^2}\alpha \)

d) \({{\cos \alpha \sin (\alpha  - 3) - \sin \alpha \cos (\alpha  - 3)} \over {\cos (3 - {\pi  \over 6}) - {1 \over 2}\sin 3}} =  - {{2\tan 3} \over {\sqrt 3 }}\)

Gợi ý làm bài

a) \(\tan 3\alpha  - \tan 2\alpha  - \tan \alpha  = \tan (2\alpha  + \alpha ) - \tan (2\alpha  + \alpha )\)

= \({{\tan 2\alpha  + \tan \alpha } \over {1 - \tan 2\alpha \tan \alpha }} - (\tan 2\alpha  + tan\alpha )\)

= \((\tan 2\alpha  + tan\alpha )({1 \over {1 - \tan 2\alpha \tan \alpha }} - 1)\)

= \(\eqalign{
& {{\tan 2\alpha + \tan \alpha } \over {1 - \tan 2\alpha \tan \alpha }}(1 - 1 + \tan 2\alpha \tan \alpha ) \cr
& = \tan 3\alpha \tan 2\alpha \tan \alpha \cr} \)

b) 

\(\eqalign{
& {{4\tan \alpha (1 - {{\tan }^2}\alpha )} \over {{{(1 + {{\tan }^2}\alpha )}^2}}} = {{2.2\tan \alpha } \over {1 + {{\tan }^2}\alpha }}.{{1 - {{\tan }^2}\alpha } \over {1 + {{\tan }^2}\alpha }} \cr
& = 2sin2\alpha c{\rm{os2}}\alpha {\rm{ = }}\sin 4\alpha \cr} \)

c) 

\(\eqalign{
& {{1 + {{\tan }^4}\alpha } \over {{{\tan }^2}\alpha + {{\cot }^2}\alpha }} = {{1 + {{\tan }^4}\alpha } \over {{{\tan }^2}\alpha + {1 \over {{{\tan }^2}\alpha }}}} \cr
& = {{1 + {{\tan }^4}\alpha } \over {{{{{\tan }^4}\alpha + 1} \over {{{\tan }^2}\alpha }}}} = {\tan ^2}\alpha \cr} \)

d) 

\(\eqalign{
& {{\cos \alpha \sin (\alpha - 3) - \sin \alpha \cos (\alpha - 3)} \over {\cos (3 - {\pi \over 6}) - {1 \over 2}\sin 3}} \cr
& = {{\sin (\alpha - 3 - \alpha )} \over {\cos 3cos{\pi \over 6} + \sin 3\sin {\pi \over 6} - {1 \over 2}\sin 3}} \cr
& = {{ - \sin 3} \over {{{\sqrt 3 } \over 2}\cos 3}} = - {{2\tan 3} \over {\sqrt 3 }} \cr} \)