Using properties of determinants,prove that:
$\left| \begin{array}{ccc} \sin \alpha & \cos \alpha & \cos (\alpha + \delta) \\ \sin \beta & \cos \beta & \cos (\beta + \delta) \\ \sin \gamma & \cos \gamma & \cos (\gamma + \delta) \end{array} \right| = 0$

(A) Let the determinant be $\Delta = \left| \begin{array}{ccc} \sin \alpha & \cos \alpha & \cos (\alpha + \delta) \\ \sin \beta & \cos \beta & \cos (\beta + \delta) \\ \sin \gamma & \cos \gamma & \cos (\gamma + \delta) \end{array} \right|$.
Using the trigonometric identity $\cos (A + B) = \cos A \cos B - \sin A \sin B$,we expand the third column:
$\Delta = \left| \begin{array}{ccc} \sin \alpha & \cos \alpha & \cos \alpha \cos \delta - \sin \alpha \sin \delta \\ \sin \beta & \cos \beta & \cos \beta \cos \delta - \sin \beta \sin \delta \\ \sin \gamma & \cos \gamma & \cos \gamma \cos \delta - \sin \gamma \sin \delta \end{array} \right|$.
Apply the column operation $C_3 \rightarrow C_3 + (\sin \delta) C_1$:
$\Delta = \left| \begin{array}{ccc} \sin \alpha & \cos \alpha & \cos \alpha \cos \delta \\ \sin \beta & \cos \beta & \cos \beta \cos \delta \\ \sin \gamma & \cos \gamma & \cos \gamma \cos \delta \end{array} \right|$.
Now,take $\cos \delta$ as a common factor from $C_3$:
$\Delta = \cos \delta \left| \begin{array}{ccc} \sin \alpha & \cos \alpha & \cos \alpha \\ \sin \beta & \cos \beta & \cos \beta \\ \sin \gamma & \cos \gamma & \cos \gamma \end{array} \right|$.
Since columns $C_2$ and $C_3$ are identical,the value of the determinant is $0$.
$\Delta = \cos \delta \times 0 = 0$.
Hence,the result is proved.