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

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

Applying $C_{1} \rightarrow+C_{1}+C_{3},$ 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 $C_{1}$ and $C_{2}$ are identical.

$\therefore \Delta=0$

Hence, the given result is proved.