Đáp án:
$\begin{array}{l}
\dfrac{{\sin a}}{{1 - \cos a}} - \dfrac{2}{{\sin a}}\\
= \dfrac{{{{\sin }^2}a - 2 + 2\cos a}}{{\sin a\left( {1 - \cos a} \right)}}\\
= \dfrac{{1 - {{\cos }^2}a - 2 + 2\cos a}}{{\sin a\left( {1 - \cos a} \right)}}\\
\left( {do:{{\sin }^2}a + {{\cos }^2}a = 1} \right)\\
= \dfrac{{\left( {1 - \cos a} \right)\left( {1 + \cos a} \right) - 2\left( {1 - \cos a} \right)}}{{\sin a\left( {1 - \cos a} \right)}}\\
= \dfrac{{\left( {1 - \cos a} \right)\left( {1 + \cos a - 2} \right)}}{{\sin a\left( {1 - \cos a} \right)}}\\
= \dfrac{{\cos a - 1}}{{\sin a}}\\
Vậy\,\dfrac{{\sin a}}{{1 - \cos a}} - \dfrac{2}{{\sin a}} = \dfrac{{\cos a - 1}}{{\sin a}}
\end{array}$
