Prove the identity: $\left(\dfrac{1 + \sin \alpha}{1 + \cos \alpha}\right) \left(\dfrac{1 + \sec \alpha}{1 + \text{cosec} \alpha}\right) = \tan \alpha$
$\begin{aligned}
LHS & = \left(\dfrac{1 + \sin \alpha}{1 + \cos \alpha}\right) \left(\dfrac{1 + \sec \alpha}{1 + \text{cosec} \alpha}\right) \\\\
& = \left(\dfrac{1 + \sin \alpha}{1 + \cos \alpha}\right) \left(\dfrac{1 + \dfrac{1}{\cos \alpha}}{1 + \dfrac{1}{\sin \alpha}}\right) \\\\
& = \left(\dfrac{1 + \sin \alpha}{1 + \cos \alpha}\right) \left[\dfrac{\left(1 + \cos \alpha\right) /\ \cos \alpha}{\left(1 + \sin \alpha\right) /\ \sin \alpha}\right] \\\\
& = \dfrac{\sin \alpha}{\cos \alpha} \\\\
& = \tan \alpha = RHS
\end{aligned}$
Hence proved.