- LG a
- LG b
- LG c
Chứng minh rằng:
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:
\[\eqalign{
& {1 \over {\sin {{10}^0}}} - {{\sqrt 3 } \over {\cos {{10}^0}}} \cr &= {{\cos {{10}^0} - \sqrt 3 \sin {{10}^0}} \over {\sin {{10}^0}\cos {{10}^0}}} \cr
& = \frac{{2.\left[ {\frac{1}{2}\cos {{10}^0} - \frac{{\sqrt 3 }}{2}\sin {{10}^0}} \right]}}{{\sin {{10}^0}\cos {{10}^0}}}\cr &= {{2[cos{{60}^0}\cos {{10}^0} - \sin {{60}^0}\sin {{10}^0}]} \over {\frac{1}{2}.2\sin {{10}^0}\cos {{10}^0}}} \cr&= {{2\cos [{{60}^0} + {{10}^0}]} \over {{1 \over 2}\sin {{20}^0}}} \cr
& = {{4\cos {{70}^0}} \over {\cos {{70}^0}}} = 4 \cr} \]