Prove the identity: $\dfrac{\sin \alpha + \sin 3 \alpha + \sin 5 \alpha}{\cos \alpha + \cos 3 \alpha + \cos 5 \alpha} = \tan 3 \alpha$
$\begin{aligned}
\sin \alpha + \sin 3 \alpha + \sin 5 \alpha & = \left(\sin \alpha + \sin 5 \alpha\right) + \sin 3 \alpha \\\\
& = 2 \sin \left(\dfrac{5 \alpha + \alpha}{2}\right) \cos \left(\dfrac{5 \alpha - \alpha}{2}\right) + \sin 3 \alpha \\\\
& = 2 \sin 3 \alpha \cos 2 \alpha + \sin 3 \alpha \\\\
& = \sin 3 \alpha \left(2 \cos 2 \alpha + 1\right) \;\;\; \cdots \; (1)
\end{aligned}$
$\begin{aligned}
\cos \alpha + \cos 3 \alpha + \cos 5 \alpha & = \left(\cos \alpha + \cos 5 \alpha\right) + \cos 3 \alpha \\\\
& = 2 \cos \left(\dfrac{5 \alpha + \alpha}{2}\right) \cos \left(\dfrac{5 \alpha - \alpha}{2}\right) + \cos 3 \alpha \\\\
& = 2 \cos 3 \alpha \cos 2 \alpha + \cos 3 \alpha \\\\
& = \cos 3 \alpha \left(2 \cos 2 \alpha + 1\right) \;\;\; \cdots \; (2)
\end{aligned}$
$\begin{aligned}
\therefore \; LHS & = \dfrac{\sin \alpha + \sin 3 \alpha + \sin 5 \alpha}{\cos \alpha + \cos 3 \alpha + \cos 5 \alpha} \\\\
& = \dfrac{\sin 3 \alpha \left(2 \cos 2 \alpha + 1\right)}{\cos 3 \alpha \left(2 \cos 2 \alpha + 1\right)} \;\;\; \left[\text{in view of equations (1) and (2)}\right] \\\\
& = \dfrac{\sin 3 \alpha}{\cos 3 \alpha} \\\\
& = \tan 3 \alpha = RHS
\end{aligned}$
Hence proved.