Bài 1)
$\begin{array}{l}
\dfrac{A}{2} + \dfrac{B}{2} + \dfrac{C}{2} = \dfrac{\pi }{2}\\
\Rightarrow \tan \left( {\dfrac{A}{2} + \dfrac{B}{2}} \right) = \tan \left( {\dfrac{\pi }{2} - \dfrac{C}{2}} \right)\\
\Rightarrow \dfrac{{\tan \dfrac{A}{2} + \tan \dfrac{B}{2}}}{{1 - \tan \dfrac{A}{2}\tan \dfrac{B}{2}}} = \cot \dfrac{C}{2} = \dfrac{1}{{\tan \dfrac{C}{2}}}\\
\Rightarrow \tan \dfrac{A}{2}\tan \dfrac{B}{2} + \tan \dfrac{B}{2}\tan \dfrac{C}{2} + \tan \dfrac{C}{2}\tan \dfrac{A}{2} = 1\\
b)\dfrac{A}{2} + \dfrac{B}{2} + \dfrac{C}{2} = \dfrac{\pi }{2}\\
\Leftrightarrow \dfrac{{\tan \dfrac{A}{2} + \tan \dfrac{B}{2}}}{{1 - \tan \dfrac{A}{2}\tan \dfrac{B}{2}}} = \cot \dfrac{C}{2}\\
\Leftrightarrow \dfrac{{\dfrac{1}{{\cot \dfrac{A}{2}}} + \dfrac{1}{{\cot \dfrac{B}{2}}}}}{{1 - \dfrac{1}{{\cot \dfrac{A}{2}\cot \dfrac{B}{2}}}}} = \cot \dfrac{C}{2}\\
\Leftrightarrow \dfrac{{\cot \dfrac{A}{2} + \cot \dfrac{B}{2}}}{{\dfrac{{\cot \dfrac{A}{2}\cot \dfrac{B}{2}}}{{\dfrac{{\cot \dfrac{A}{2}\cot \dfrac{B}{2} - 1}}{{\cot \dfrac{A}{2}\cot \dfrac{B}{2}}}}}}} = \cot \dfrac{C}{2}\\
\Leftrightarrow \dfrac{{\cot \dfrac{A}{2} + \cot \dfrac{B}{2}}}{{\cot \dfrac{A}{2}\cot \dfrac{B}{2} - 1}} = \cot \dfrac{C}{2}\\
\Leftrightarrow \cot \dfrac{A}{2} + \cot \dfrac{B}{2} + \cot \dfrac{C}{2} = \cot \dfrac{A}{2}\cot \dfrac{B}{2}\cot \dfrac{C}{2}
\end{array}$
Bài 2:
$\begin{array}{l}
a)\cos \left( {a + b} \right) = 2\cos \left( {a - b} \right)\\
\Leftrightarrow \left( {\cos a\cos b - \sin a\sin b} \right) = 2\left( {\cos a\cos b + \sin a\sin b} \right)\\
\Leftrightarrow - \cos a\cos b = 3\sin a\sin b\\
\Leftrightarrow - \dfrac{1}{3} = \dfrac{{\sin a\sin b}}{{\cos a\cos b}} = \tan a\tan b\\
b)\cos \left( {a + 2b} \right) = k\cos a\\
\Leftrightarrow \cos \left( {a + b + b} \right) = k\cos \left( {a + b - b} \right)\\
\Leftrightarrow \cos \left( {a + b} \right)\cos b - \sin \left( {a + b} \right)\sin b = k\cos \left( {a + b} \right)\cos b + k\sin b\sin \left( {a + b} \right)\\
\Leftrightarrow \left( {1 + k} \right)\sin b\sin \left( {a + b} \right) = \cos b\cos \left( {a + b} \right)\left( {1 - k} \right)\\
\Leftrightarrow \dfrac{{\sin b\sin \left( {a + b} \right)}}{{\cos b\cos \left( {a + b} \right)}} = \dfrac{{1 - k}}{{1 + k}}\\
\Leftrightarrow \tan \left( {a + b} \right)\tan b = \dfrac{{1 - k}}{{1 + k}}
\end{array}$
