Solution:

Given:

$ \begin{aligned} a &=\cos \alpha+i \sin \alpha \\\\ b &=\cos \beta+i \sin \beta \\\\ \& c & =\cos \gamma+i \sin \gamma \end{aligned}\\ $

Now,

$ \frac{\mathrm{ab}}{\mathrm{c}}=\mathrm{ab}(\mathrm{c})^{-1}\\ $

$ =(\cos \alpha+i \sin \alpha)(\cos \beta+i \sin \beta)(\cos \gamma+i \sin \gamma)^{-1}\\ $

$ =(\cos \alpha+i \sin \alpha)(\cos \beta+i \sin \beta)(\cos \gamma-i \sin \gamma)\\ $

$ \Rightarrow \frac{a b}{c}=\cos (\alpha+\beta-\gamma)+i \sin (\alpha+\beta-\gamma)\\ $

$ also, \frac{c}{a b}=\left[[\frac{a b}{c}]^{-1}\right]\\ $

$ =[\cos (\alpha+\beta-\gamma)+i \sin (\alpha+\beta-\gamma)]^{-1}\\ $

$ \Rightarrow \frac{\mathrm{c}}{\mathrm{ab}}=\cos (\alpha+\beta-\gamma)-i \sin (\alpha+\beta-\gamma)\\ $

$$\therefore(1)-(2)\ then, $$

$\frac{\mathrm{ab}}{\mathrm{c}}-\frac{\mathrm{c}}{\mathrm{ab}}=\cos (\alpha+\beta-\gamma)+i \sin (\alpha+\beta-\gamma)-\cos (\alpha+\beta-\gamma)+i \sin (\alpha+\beta-\gamma)\\$ $\Rightarrow \frac{\mathrm{ab}}{\mathrm{c}}-\frac{\mathrm{c}}{\mathrm{ab}}=2 \mathrm{i} \sin (\alpha+\beta-\gamma)$