Prove the identity: $\dfrac{\cot \alpha + \tan \alpha}{1 + \tan 2 \alpha \tan \alpha} = 2 \cot 2 \alpha$
$\begin{aligned}
\cot \alpha + \tan \alpha & = \dfrac{\cos \alpha}{\sin \alpha} + \dfrac{\sin \alpha}{\cos \alpha} \\\\
& = \dfrac{\cos^2 \alpha + \sin^2 \alpha}{\sin \alpha \cos \alpha} \\\\
& = \dfrac{1}{\sin \alpha \cos \alpha} \\\\
& = \dfrac{2}{2 \sin \alpha \cos \alpha} \\\\
& = \dfrac{2}{\sin 2 \alpha} \;\;\; \cdots \; (1)
\end{aligned}$
$\begin{aligned}
1 + \tan 2 \alpha \tan \alpha & = 1 + \left(\dfrac{2 \tan \alpha}{1 - \tan^2 \alpha}\right) \tan \alpha \\\\
& = \dfrac{1 - \tan^2 \alpha + 2 \tan^2 \alpha}{1 - \tan^2 \alpha} \\\\
& = \dfrac{1 + \tan^2 \alpha}{1 - \tan^2 \alpha} \\\\
& = \dfrac{1 + \dfrac{\sin^2 \alpha}{\cos^2 \alpha}}{1 - \dfrac{\sin^2 \alpha}{\cos^2 \alpha}} \\\\
& = \dfrac{\cos^2 \alpha + \sin^2 \alpha}{\cos^2 \alpha - \sin^2 \alpha} \\\\
& = \dfrac{1}{\cos^2 \alpha - \sin^2 \alpha} \\\\
& = \dfrac{1}{\cos 2 \alpha} \;\;\; \cdots \; (2)
\end{aligned}$
$\begin{aligned}
\therefore \; LHS & = \dfrac{\cot \alpha + \tan \alpha}{1 + \tan 2 \alpha \tan \alpha} \\\\
& = \dfrac{\dfrac{2}{\sin 2 \alpha}}{\dfrac{1}{\cos 2 \alpha}} \;\;\; \left[\text{By equations (1) and (2)}\right] \\\\
& = 2 \times \dfrac{\cos 2 \alpha}{\sin 2 \alpha} \\\\
& = 2 \cot 2 \alpha = RHS
\end{aligned}$
Hence proved.