Bài tập 52 trang 216 SGK Toán 10 NC

a) Ta có:

\(\begin{array}{*{20}{l}}
{\tan \alpha  + \tan \beta  = \frac{{\sin \alpha }}{{\cos \alpha }} + \frac{{\sin \beta }}{{\cos \beta }}}\\
\begin{array}{l}
 = \frac{{\sin \alpha \cos \beta  + \sin \beta \cos \alpha }}{{\cos \alpha \cos \beta }}\\
 = \frac{{\sin \left( {\alpha  + \beta } \right)}}{{\cos \alpha \cos \beta }}
\end{array}
\end{array}\)

Tương tự, 

\(\tan \alpha  - \tan \beta  = \frac{{\sin \left( {\alpha  - \beta } \right)}}{{\cos \alpha \cos \beta }}\)

b)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\frac{1}{{\cos \alpha \cos 2\alpha }} + \frac{1}{{\cos 2\alpha \cos 3\alpha }}\\
 + ... + \frac{1}{{\cos 7\alpha \cos 8\alpha }}
\end{array}\\
\begin{array}{l}
 = \frac{{\tan \alpha  - \tan 2\alpha }}{{\sin \left( { - \alpha } \right)}} + \frac{{\tan 2\alpha  - \tan 3\alpha }}{{\sin \left( { - \alpha } \right)}}\\
 + ... + \frac{{\tan 7\alpha  - \tan 8\alpha }}{{\sin \left( { - \alpha } \right)}}
\end{array}\\
\begin{array}{l}
 = \frac{1}{{\sin \alpha }}( - \tan \alpha  + \tan 2\alpha  - \tan 2\alpha \\
 + \tan 3\alpha  + ... - \tan 7\alpha  + \tan 8\alpha )
\end{array}\\
{ = \frac{{\tan 8\alpha  - \tan \alpha }}{{\sin \alpha }}}
\end{array}\)

-- Mod Toán 10 HỌC247