Bài 58 trang 218 SGK Đại số 10 Nâng cao

Đề bài

Chứng minh rằng:

a) Nếu \(α + β + γ = kπ (k ∈ Z)\) và \(\cosα \cosβ \cosγ ≠ 0\) thì

\( \tan \alpha + \tan \beta + \tan \gamma = \tan \alpha \tan \beta \tan \gamma\)

b) Nếu \(0 < \alpha  < \beta  < \gamma  < {\pi  \over 2}\) và \(\tan \alpha  = {1 \over 8};\,\tan \beta  = {1 \over 5};\,\tan \gamma  = {1 \over 2}\) thì \(\alpha  + \beta  + \gamma  = {\pi  \over 2}\)

c) \({1 \over {\sin {{10}^0}}} - {{\sqrt 3 } \over {\cos {{10}^0}}} = 4\)

Hướng dẫn giải

a) Ta có: \(α + β +  γ = kπ  \)

\(⇒ tan (α + β ) = tan(kπ – γ) = - tanγ\)

\(\eqalign{
& \Rightarrow {{\tan \alpha + \tan \gamma } \over {1 - \tan \alpha \tan \beta }} = - \tan \gamma\cr& \Rightarrow \tan \alpha + \tan \beta = - \tan \gamma (1 - \tan \alpha \tan \beta ) \cr
& \Rightarrow \tan \alpha + \tan \beta + \tan \gamma = \tan \alpha \tan \beta \tan \gamma \cr} \) 

b) Ta có:

\(\eqalign{
& \tan (\alpha + \beta ) = {{\tan \alpha + \tan \beta } \over {1 - \tan \alpha \tan \beta }} = {{{1 \over 8} + {1 \over 5}} \over {1 - {1 \over 8}.{1 \over 5}}} = {1 \over 3} \cr
& \Rightarrow \tan (\alpha + \beta + \gamma ) = {{\tan (\alpha + \beta ) + \tan \gamma } \over {1 - \tan (\alpha + \beta )  \tan \gamma }} \cr&= {{{1 \over 3} + {1 \over 2}} \over {1 - {1 \over 3}.{1 \over 2}}} = 1 \cr} \) 

Vì \(0 < \alpha  + \beta  + \gamma  < {{3\pi } \over 2} \Rightarrow \alpha  + \beta  + \gamma  = {\pi  \over 4}\)

c) Ta có:

\(\eqalign{
& {1 \over {\sin {{10}^0}}} - {{\sqrt 3 } \over {\cos {{10}^0}}} = {{\cos {{10}^0} - \sqrt 3 \sin {{10}^0}} \over {\sin {{10}^0}\cos {{10}^0}}} \cr
& = {{2(cos{{60}^0}\cos {{10}^0} - \sin {{60}^0}\sin {{10}^0})} \over {\sin {{10}^0}\cos {{10}^0}}} \cr&= {{2\cos ({{60}^0} + {{10}^0})} \over {{1 \over 2}\sin {{20}^0}}} \cr
& = {{4\cos {{70}^0}} \over {\cos {{70}^0}}} = 4 \cr} \)