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
Đáp án
a) Ta có: α + β + γ = kπ
⇒ tan (α + β ) = tan(kπ – γ) = - tanγ
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\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}