Biến đổi \[\tan \alpha + \tan \beta \]

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

Cũng có thể áp dụng công thức cộng:
\[\cos \alpha \cos \beta  = \frac{1}{2}\left[ {c{\rm{os}}\left( {\alpha  - \beta } \right) + c{\rm{os}}\left( {\alpha  + \beta } \right)} \right]\]

\[\begin{array}{l}
  \Rightarrow \frac{{\sin \left( {\alpha  + \beta } \right)}}{{\cos \alpha \cos \beta }} = \frac{{\sin \left( {\alpha  + \beta } \right)}}{{\frac{1}{2}\left[ {c{\rm{os}}\left( {\alpha  - \beta } \right) + c{\rm{os}}\left( {\alpha  + \beta } \right)} \right]}} \\
  = \frac{{2\sin \left( {\alpha  + \beta } \right)}}{{c{\rm{os}}\left( {\alpha  - \beta } \right) + c{\rm{os}}\left( {\alpha  + \beta } \right)}} \\
 \end{array}\]

Vậy:

\[\tan \alpha  + \tan \beta  = \frac{{\sin \left( {\alpha  + \beta } \right)}}{{\cos \alpha \cos \beta }} = \frac{{2\sin \left( {\alpha  + \beta } \right)}}{{c{\rm{os}}\left( {\alpha  - \beta } \right) + c{\rm{os}}\left( {\alpha  + \beta } \right)}}\]