Prove the identity: $\dfrac{\sin \left(\pi - \alpha\right)}{\sin \alpha - \cos \alpha \tan \dfrac{\alpha}{2}} + \cos \left(\pi - \alpha\right) = 1$
$\begin{aligned}
LHS & = \dfrac{\sin \left(\pi - \alpha\right)}{\sin \alpha - \cos \alpha \tan \dfrac{\alpha}{2}} + \cos \left(\pi - \alpha\right) \\\\
& = \dfrac{\sin \alpha}{\sin \alpha - \cos \alpha \left(\dfrac{1 - \cos \alpha}{\sin \alpha}\right)} - \cos \alpha \\\\
& = \dfrac{\sin^2 \alpha}{\sin^2 \alpha - \cos \alpha + \cos^2 \alpha} - \cos \alpha \\\\
& = \dfrac{\sin^2 \alpha}{1 - \cos \alpha} - \cos \alpha \\\\
& = \dfrac{\sin^2 \alpha - \cos \alpha + \cos^2 \alpha}{1 - \cos \alpha} \\\\
& = \dfrac{1 - \cos \alpha}{1 - \cos \alpha} \\\\
& = 1 = RHS
\end{aligned}$
Hence proved.