Bài tập 58 trang 218 SGK Toán 10 NC

a) Ta có:

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\alpha  + \beta  + \gamma  = k\pi \\
 \Rightarrow \tan \left( {\alpha  + \beta } \right) = \tan \left( {k\pi  - \gamma } \right) =  - \tan \gamma 
\end{array}\\
{ \Rightarrow \frac{{\tan \alpha  + \tan \beta }}{{1 - \tan \alpha \tan \beta }} =  - \tan \gamma }\\
{ \Rightarrow \tan \alpha  + \tan \beta  =  - \tan \gamma \left( {1 - \tan \alpha \tan \beta } \right)}\\
{ \Rightarrow \tan \alpha  + \tan \beta  + \tan \gamma  = \tan \alpha \tan \beta \tan \gamma }
\end{array}\)

b)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\tan \left( {\alpha  + \beta } \right) = \frac{{\tan \alpha  + \tan \beta }}{{1 - \tan \alpha \tan \beta }}\\
 = \frac{{\frac{1}{8} + \frac{1}{5}}}{{1 - \frac{1}{8}.\frac{1}{5}}} = \frac{1}{3}
\end{array}\\
\begin{array}{l}
 \Rightarrow \tan \left( {\alpha  + \beta  + \gamma } \right)\\
 = \frac{{\tan \left( {\alpha  + \beta } \right) + \tan \gamma }}{{1 - \tan \left( {\alpha  + \beta } \right)\tan \lambda }} = \frac{{\frac{1}{3} + \frac{1}{2}}}{{1 - \frac{1}{3}.\frac{1}{2}}} = 1
\end{array}
\end{array}\)

Vì \(0 < \alpha  + \beta  + \gamma  < \frac{{3\pi }}{2}\) nên ta có \(\alpha  + \beta  + \gamma  = \frac{\pi }{4}\)

c)

\(\begin{array}{l}
\frac{1}{{\sin {{10}^0}}} - \frac{{\sqrt 3 }}{{\cos {{10}^0}}} = \frac{{\cos {{10}^0} - \sqrt 3 \sin {{10}^0}}}{{\sin {{10}^0}\cos {{10}^0}}}\\
 = \frac{{2\left( {\cos {{60}^0}\cos {{10}^0} - \sin {{60}^0}\sin {{10}^0}} \right)}}{{\sin {{10}^0}\cos {{10}^0}}}\\
 = \frac{{2\cos \left( {{{60}^0} + {{10}^0}} \right)}}{{\frac{1}{2}\sin {{20}^0}}} = \frac{{4\cos {{70}^0}}}{{\cos {{70}^0}}} = 4
\end{array}\)

-- Mod Toán 10 HỌC247