Using properties of determinants, prove that:

$\Delta=\left|\begin{array}{lll}\sin \alpha & \cos \alpha & \cos (\alpha+\delta) \\ \sin \beta & \cos \beta & \cos (\beta+\delta) \\ \sin \gamma & \cos \gamma & \cos (\gamma+\delta)\end{array}\right|$

$=\frac{1}{\sin \delta \cos \delta}$$\left|\begin{array}{lll}\sin \alpha \sin \delta & \cos \alpha \cos \delta & \cos \alpha \cos \delta-\sin \alpha \sin \delta \\ \sin \beta \sin \delta & \cos \beta \cos \delta & \cos \beta \cos \delta-\sin \beta \sin \delta \\ \sin \gamma \sin \delta & \cos \gamma \cos \delta & \cos \gamma \cos \delta-\sin \gamma \sin \delta\end{array}\right|$

Applying $\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{3}$, we have:

$\Delta=\frac{1}{\sin \delta \cos \delta}\left|\begin{array}{lll}\cos \alpha \cos \delta & \cos \alpha \cos \delta & \cos \alpha \cos \delta-\sin \alpha \sin \delta \\ \cos \beta \cos \delta & \cos \beta \cos \delta & \cos \beta \cos \delta-\sin \beta \sin \delta \\ \cos \gamma \cos \delta & \cos \gamma \cos \delta & \cos \gamma \cos \delta-\sin \gamma \sin \delta\end{array}\right|$

Here, two columns $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ are identical.

$\therefore \Delta=0$

Hence, the given result is proved.