Prove the identity: $\dfrac{\tan 3 \alpha}{\tan \alpha} = \dfrac{3 - \tan^2 \alpha}{1 - 3 \tan^2 \alpha}$
$\begin{aligned}
LHS & = \dfrac{\tan 3 \alpha}{\tan \alpha} \\\\
& = \dfrac{\tan \left(2 \alpha + \alpha\right)}{\tan \alpha} \\\\
& = \left(\dfrac{\tan 2 \alpha + \tan \alpha}{1 - \tan 2 \alpha \; \tan \alpha}\right) \times \dfrac{1}{\tan \alpha} \\\\
& = \left(\dfrac{\dfrac{2 \tan \alpha}{1 - \tan^2 \alpha} + \tan \alpha}{1 - \dfrac{2 \tan \alpha}{1 - \tan^2 \alpha}\times \tan \alpha}\right) \times \dfrac{1}{\tan \alpha} \\\\
& = \left(\dfrac{2 \tan \alpha + \tan \alpha - \tan^3 \alpha}{1 - \tan^2 \alpha - 2 \tan^2 \alpha}\right) \times \dfrac{1}{\tan \alpha} \\\\
& = \left(\dfrac{3 \tan \alpha - \tan^3 \alpha}{1 - 3 \tan^2 \alpha}\right) \times \dfrac{1}{\tan \alpha} \\\\
& = \dfrac{3 - \tan^2 \alpha}{1 - 3 \tan^2 \alpha} = RHS
\end{aligned}$
Hence proved.