If $\;$ $a = \cos 2 \alpha + i \sin 2 \alpha$, $\;$ $b = \cos 2 \beta + i \sin 2 \beta$ $\;$ and $\;$ $c = \cos 2 \gamma + i \sin 2 \gamma$, $\;$ prove that $\;$ $\dfrac{a^2 b^2 + c^2}{abc} = 2 \cos 2 \left(\alpha + \beta - \gamma\right)$
Given: $\;$ $a = \cos 2 \alpha + i \sin 2 \alpha$, $\;$ $b = \cos 2 \beta + i \sin 2 \beta$, $\;$ $c = \cos 2 \gamma + i \sin 2 \gamma$
$\therefore$ $\;$ $a^2 = \left(\cos 2 \alpha + i \sin 2 \alpha\right)^2 = \cos 4 \alpha + i \sin 4 \alpha$
$b^2 = \left(\cos 2 \beta + i \sin 2 \beta\right)^2 = \cos 4 \beta + i \sin 4 \beta$
$c^2 = \left(\cos 2 \gamma + i \sin 2 \gamma\right)^2 = \cos 4 \gamma + i \sin 4 \gamma$
Now,
$\begin{aligned}
a^2 b^2 & = \left(\cos 4 \alpha + i \sin 4 \alpha\right) \left(\cos 4 \beta + i \sin 4 \beta\right) \\\\
& = \left(\cos 4 \alpha \; \cos 4 \beta - \sin 4 \alpha \; \sin 4 \beta\right) + i \left(\sin 4 \alpha \; \cos 4 \beta + \cos 4 \alpha \; \sin 4 \beta\right) \\\\
& = \cos \left(4 \alpha + 4 \beta\right) + i \sin \left(4 \alpha + 4 \beta\right)
\end{aligned}$
$\begin{aligned}
\therefore \; a^2 b^2 + c^2 & = \left[\cos \left(4 \alpha + 4 \beta\right) + i \sin \left(4 \alpha + 4 \beta\right)\right] + \left[\cos 4 \gamma + i \sin 4 \gamma\right] \\\\
& = \left[\cos \left(4 \alpha + 4 \beta\right) + \cos 4 \gamma\right] + i \left[\sin \left(4 \alpha + 4 \beta\right) + \sin 4 \gamma\right] \\\\
& = 2 \cos \left(\dfrac{4 \alpha + 4 \beta + 4 \gamma}{2}\right) \cos \left(\dfrac{4 \alpha + 4 \beta - 4 \gamma}{2}\right) \\
& \hspace{1cm} + 2 i \sin \left(\dfrac{4 \alpha + 4 \beta + 4 \gamma}{2}\right) \cos \left(\dfrac{4 \alpha + 4 \beta - 4 \gamma}{2}\right) \\\\
& = 2 \cos \left(2 \alpha + 2 \beta - 2 \gamma\right)\left[\cos \left(2 \alpha + 2 \beta + 2 \gamma\right) + i \sin \left(2 \alpha + 2 \beta + 2 \gamma\right)\right]
\end{aligned}$
$\begin{aligned}
abc & = \left(\cos 2 \alpha + i \sin 2 \alpha\right) \left(\cos 2 \beta + i \sin 2 \beta\right) \left(\cos 2 \gamma + i \sin 2 \gamma\right) \\\\
& = \left[\left(\cos 2 \alpha \; \cos 2 \beta - \sin 2 \alpha \; \sin 2 \beta \right) + i \left(\sin 2 \alpha \; \cos 2 \beta + \cos 2 \alpha \; \sin 2 \beta \right)\right] \\
& \hspace{9cm} \times \left(\cos 2 \gamma + i \sin 2 \gamma\right) \\\\
& = \left[\cos \left(2 \alpha + 2 \beta\right) + i \sin \left(2 \alpha + 2 \beta\right)\right] \times \left(\cos 2 \gamma + i \sin 2 \gamma\right) \\\\
& = \left[\cos \left(2 \alpha + 2 \beta\right) \; \cos 2 \gamma - \sin \left(2 \alpha + 2 \beta\right) \; \sin 2 \gamma\right] \\
& \hspace{1cm} + i \left[\sin \left(2 \alpha + 2 \beta\right) \; \cos 2 \gamma + \cos \left(2 \alpha + 2 \beta \right) \; \sin 2 \gamma\right] \\\\
& = \cos \left(2 \alpha + 2 \beta + 2 \gamma\right) + i \sin \left(2 \alpha + 2 \beta + 2 \gamma\right)
\end{aligned}$
$\begin{aligned}
\therefore \; \dfrac{a^2 b^2 + c^2}{abc} & = \dfrac{2 \cos \left(2 \alpha + 2 \beta - 2 \gamma\right)\left[\cos \left(2 \alpha + 2 \beta + 2 \gamma\right) + i \sin \left(2 \alpha + 2 \beta + 2 \gamma\right)\right]}{\cos \left(2 \alpha + 2 \beta + 2 \gamma\right) + i \sin \left(2 \alpha + 2 \beta + 2 \gamma\right)} \\\\
& = 2 \cos \left(2 \alpha + 2 \beta - 2 \gamma\right) \\\\
& = 2 \cos 2 \left(\alpha + \beta - \gamma\right)
\end{aligned}$
Hence proved.