Đáp án:
$\begin{array}{l}
\tan \left( {a + b} \right)\\
= \dfrac{{\sin \left( {a + b} \right)}}{{\cos \left( {a + b} \right)}}\\
= \dfrac{{\sin a.\cos b + \sin b.\cos a}}{{\cos a.\cos b - \sin a.\sin b}}\\
\left( {Do:\left\{ \begin{array}{l}
\sin \left( {a + b} \right) = \sin a.\cos b + \sin b.\cos a\\
\cos \left( {a + b} \right) = \cos a.\cos b - \sin a.\sin b
\end{array} \right.} \right)\\
= \dfrac{{\dfrac{{\sin a.\cos b}}{{\cos a.\cos b}} + \dfrac{{\sin b.\cos a}}{{\cos a.\cos b}}}}{{1 - \dfrac{{\sin a.\sin b}}{{\cos a.\cos b}}}}\\
= \dfrac{{\dfrac{{\sin a}}{{\cos a}} + \dfrac{{\sin b}}{{\cos b}}}}{{1 - \dfrac{{\sin a}}{{\cos a}}.\dfrac{{\sin b}}{{\cos b}}}}\\
= \dfrac{{\tan a + \tan b}}{{1 - \tan a.\tan b}}\\
Vậy\,\tan \left( {a + b} \right) = \dfrac{{\tan + \tan b}}{{1 - \tan a.\tan b}}
\end{array}$
