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Câu hỏi: Chứng minh rằng:
\( \tan \alpha + \tan \beta + \tan \gamma = \tan \alpha \tan \beta \tan \gamma\)
Lời giải chi tiết:
Ta có: \(α + β + γ = kπ \)
\(\begin{array}{l}
\Rightarrow \alpha + \beta = k\pi - \gamma \\
\Rightarrow \tan \left({\alpha + \beta } \right) = \tan \left({k\pi - \gamma } \right)\\
\Rightarrow \tan \left({\alpha + \beta } \right) = \tan \left({ - \gamma } \right)\\
\Rightarrow \tan \left({\alpha + \beta } \right) = - \tan \gamma
\end{array}\)
\(\eqalign{
& \Rightarrow {{\tan \alpha + \tan \beta } \over {1 - \tan \alpha \tan \beta }} = - \tan \gamma\cr& \Rightarrow \tan \alpha + \tan \beta = - \tan \gamma (1 - \tan \alpha \tan \beta) \cr
& \Rightarrow \tan \alpha + \tan \beta = - \tan \gamma + \tan \alpha \tan \beta \tan \gamma \cr&\Rightarrow \tan \alpha + \tan \beta + \tan \gamma = \tan \alpha \tan \beta \tan \gamma \cr} \)
Lời giải chi tiết:
Ta có:
\(\eqalign{
& \tan (\alpha + \beta) = {{\tan \alpha + \tan \beta } \over {1 - \tan \alpha \tan \beta }}\cr & = {{{1 \over 8} + {1 \over 5}} \over {1 - {1 \over 8}.{1 \over 5}}} = {1 \over 3} \cr
& \Rightarrow \tan (\alpha + \beta + \gamma) \cr &= {{\tan (\alpha + \beta) + \tan \gamma } \over {1 - \tan (\alpha + \beta) \tan \gamma }} \cr&= {{{1 \over 3} + {1 \over 2}} \over {1 - {1 \over 3}.{1 \over 2}}} = 1 \cr} \)
Vì \(0 < \alpha + \beta + \gamma < {{3\pi } \over 2} \) \(\Rightarrow \alpha + \beta + \gamma = {\pi \over 4}\)
Lời giải chi tiết:
Ta có:
\(\eqalign{
& {1 \over {\sin {{10}^0}}} - {{\sqrt 3 } \over {\cos {{10}^0}}} \cr &= {{\cos {{10}^0} - \sqrt 3 \sin {{10}^0}} \over {\sin {{10}^0}\cos {{10}^0}}} \cr
& = \frac{{2.\left({\frac{1}{2}\cos {{10}^0} - \frac{{\sqrt 3 }}{2}\sin {{10}^0}} \right)}}{{\sin {{10}^0}\cos {{10}^0}}}\cr &= {{2(cos{{60}^0}\cos {{10}^0} - \sin {{60}^0}\sin {{10}^0})} \over {\frac{1}{2}. 2\sin {{10}^0}\cos {{10}^0}}} \cr&= {{2\cos ({{60}^0} + {{10}^0})} \over {{1 \over 2}\sin {{20}^0}}} \cr
& = {{4\cos {{70}^0}} \over {\cos {{70}^0}}} = 4 \cr} \)
Câu a
Nếu \(α + β + γ = kπ (k ∈ Z)\) và \(\cosα \cosβ \cosγ ≠ 0\) thì\( \tan \alpha + \tan \beta + \tan \gamma = \tan \alpha \tan \beta \tan \gamma\)
Lời giải chi tiết:
Ta có: \(α + β + γ = kπ \)
\(\begin{array}{l}
\Rightarrow \alpha + \beta = k\pi - \gamma \\
\Rightarrow \tan \left({\alpha + \beta } \right) = \tan \left({k\pi - \gamma } \right)\\
\Rightarrow \tan \left({\alpha + \beta } \right) = \tan \left({ - \gamma } \right)\\
\Rightarrow \tan \left({\alpha + \beta } \right) = - \tan \gamma
\end{array}\)
\(\eqalign{
& \Rightarrow {{\tan \alpha + \tan \beta } \over {1 - \tan \alpha \tan \beta }} = - \tan \gamma\cr& \Rightarrow \tan \alpha + \tan \beta = - \tan \gamma (1 - \tan \alpha \tan \beta) \cr
& \Rightarrow \tan \alpha + \tan \beta = - \tan \gamma + \tan \alpha \tan \beta \tan \gamma \cr&\Rightarrow \tan \alpha + \tan \beta + \tan \gamma = \tan \alpha \tan \beta \tan \gamma \cr} \)
Câu 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 \(\alpha + \beta + \gamma = {\pi \over 4}\)Lời giải chi tiết:
Ta có:
\(\eqalign{
& \tan (\alpha + \beta) = {{\tan \alpha + \tan \beta } \over {1 - \tan \alpha \tan \beta }}\cr & = {{{1 \over 8} + {1 \over 5}} \over {1 - {1 \over 8}.{1 \over 5}}} = {1 \over 3} \cr
& \Rightarrow \tan (\alpha + \beta + \gamma) \cr &= {{\tan (\alpha + \beta) + \tan \gamma } \over {1 - \tan (\alpha + \beta) \tan \gamma }} \cr&= {{{1 \over 3} + {1 \over 2}} \over {1 - {1 \over 3}.{1 \over 2}}} = 1 \cr} \)
Vì \(0 < \alpha + \beta + \gamma < {{3\pi } \over 2} \) \(\Rightarrow \alpha + \beta + \gamma = {\pi \over 4}\)
Câu c
\({1 \over {\sin {{10}^0}}} - {{\sqrt 3 } \over {\cos {{10}^0}}} = 4\)Lời giải chi tiết:
Ta có:
\(\eqalign{
& {1 \over {\sin {{10}^0}}} - {{\sqrt 3 } \over {\cos {{10}^0}}} \cr &= {{\cos {{10}^0} - \sqrt 3 \sin {{10}^0}} \over {\sin {{10}^0}\cos {{10}^0}}} \cr
& = \frac{{2.\left({\frac{1}{2}\cos {{10}^0} - \frac{{\sqrt 3 }}{2}\sin {{10}^0}} \right)}}{{\sin {{10}^0}\cos {{10}^0}}}\cr &= {{2(cos{{60}^0}\cos {{10}^0} - \sin {{60}^0}\sin {{10}^0})} \over {\frac{1}{2}. 2\sin {{10}^0}\cos {{10}^0}}} \cr&= {{2\cos ({{60}^0} + {{10}^0})} \over {{1 \over 2}\sin {{20}^0}}} \cr
& = {{4\cos {{70}^0}} \over {\cos {{70}^0}}} = 4 \cr} \)
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