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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: \(α + β + γ = kπ \)
\(⇒ tan (α + β ) = tan(kπ – γ) = – tanγ\)
\(\eqalign{
& \Rightarrow {{\tan \alpha + \tan \gamma } \over {1 – \tan \alpha \tan \beta }} = – \tan \gamma\cr& \Rightarrow \tan \alpha + \tan \beta = – \tan \gamma (1 – \tan \alpha \tan \beta ) \cr
& \Rightarrow \tan \alpha + \tan \beta + \tan \gamma = \tan \alpha \tan \beta \tan \gamma \cr} \)
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:
\(\eqalign{
& {1 \over {\sin {{10}^0}}} – {{\sqrt 3 } \over {\cos {{10}^0}}} = {{\cos {{10}^0} – \sqrt 3 \sin {{10}^0}} \over {\sin {{10}^0}\cos {{10}^0}}} \cr
& = {{2(cos{{60}^0}\cos {{10}^0} – \sin {{60}^0}\sin {{10}^0})} \over {\sin {{10}^0}\cos {{10}^0}}} \cr&= {{2\cos ({{60}^0} + {{10}^0})} \over {{1 \over 2}\sin {{20}^0}}} \cr
& = {{4\cos {{70}^0}} \over {\cos {{70}^0}}} = 4 \cr} \)