Without expanding, show that the value of each of the determinants is zero: `|sinalphacosalpha"cos"(alpha+delta)sinbetacosbeta"cos"(beta+delta)singammacosgamma"cos"(gamma+delta)|`

Explanation

$$\Delta=\left|\begin{array}{lll}\sin \alpha & \cos \alpha & \cos (\alpha+\delta) \\ \sin \beta & \cos \beta & \cos (\beta+\delta) \\ \sin \gamma & \cos \gamma & \cos (\gamma+\delta)\end{array}\right|$$
Multiplying `C_{1}` with `sin delta, C_{2} `with `cos delta,` we obtain
$$
\Rightarrow \Delta=\frac{1}{\sin \delta \cos \delta}\left|\begin{array}{lll}
\sin \alpha \sin \delta & \cos \alpha \cos \delta & \cos (\alpha+\delta) \\
\sin \beta \sin \delta & \cos \beta \cos \delta & \cos (\beta+\delta) \\
\sin \gamma \sin \delta & \cos \gamma \cos \delta & \cos (\gamma+\delta)
\end{array}\right|
...