\(\left|\begin{array}{ccc}\sin \alpha & \cos \alpha & \sin (\alpha+\delta) \\ \sin \beta & \cos \beta & \sin (\beta+\delta) \\ \sin \gamma & \cos \gamma & \sin (\gamma+\delta)\end{array}\right|\) is equal to

Given, \(\left|\begin{array}{ccc}\sin \alpha & \cos \alpha & \sin (\alpha+\delta) \\ \sin \beta & \cos \beta & \sin (\beta+\delta) \\ \sin \gamma & \cos \gamma & \sin (\gamma+\delta)\end{array}\right|\)
\(=\left|\begin{array}{ccc}\sin \alpha & \cos \alpha & \sin \alpha \cdot \cos \delta+\cos \alpha \cdot \sin \delta \\ \sin \beta & \cos \beta & \sin \beta \cdot \cos \delta+\cos \beta \cdot \sin \delta \\ \sin \gamma & \cos \gamma & \sin \gamma \cdot \cos \delta+\cos \gamma \cdot \sin \delta\end{array}\right|\)
\(=\left|\begin{array}{ccc}\sin \alpha & \cos \alpha & \sin \alpha \cdot \cos \delta \\ \sin \beta & \cos \beta & \sin \beta \cdot \cos \delta \\ \sin \gamma & \cos \gamma & \sin \gamma \cdot \cos \delta\end{array}\right|\)
\(+\left|\begin{array}{ccc}
\sin \alpha & \cos \alpha & \cos \alpha \cdot \sin \delta \\
\sin \beta & \cos \beta & \cos \beta \cdot \sin \delta \\
\sin \gamma & \cos \gamma & \cos \gamma \cdot \sin \delta
\end{array}\right|\)
\(=\cos \delta\left|\begin{array}{ccc}\sin \alpha & \cos \alpha & \sin \alpha \\ \sin \beta & \cos \beta & \sin \beta \\ \sin \gamma & \cos \gamma & \sin \gamma\end{array}\right|\)
\(+\sin \delta\left|\begin{array}{ccc}\sin \alpha & \cos \alpha & \cos \alpha \\ \sin \beta & \cos \beta & \cos \beta \\ \sin \gamma & \cos \gamma & \cos \gamma\end{array}\right|\)
\[
=\cos \delta \times 0+\sin \delta \times 0
\]
\(\left(\because C_{1}\right.\) and \(C_{2}\) are identical \(=0\) \(C_{2}\) and \(C_{3}\) identical.)