\[\frac{\left( 1 - \cos\alpha \right)\left( 1 + \cos\alpha \right)}{\sin\alpha} =\]
\[= \frac{1 - \cos^{2}\alpha}{\sin a} = \frac{\sin^{2}a}{\sin a} = \sin a;\ \ \ \ \ \ \ \]
\[a \neq \pi k.\]