Bài 58 trang 218 SGK Đại số 10 Nâng cao

LG 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: \(α + β +  γ = kπ  \)

\(\begin{array}{l}
\Rightarrow \alpha + \beta = k\pi - \gamma \\
\Rightarrow \tan \left( {\alpha + \beta } \right) = \tan \left( {k\pi - \gamma } \right)\\
\Rightarrow \tan \left( {\alpha + \beta } \right) = \tan \left( { - \gamma } \right)\\
\Rightarrow \tan \left( {\alpha + \beta } \right) = - \tan \gamma
\end{array}\)

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

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

LG 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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