. $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{\sin \alpha }&{\cos \alpha }&{\cos (\alpha + \delta )}\\{\sin \beta }&{\cos \beta }&{\cos (\beta + \delta )}\\{\sin \gamma }&{\cos \gamma }&{\cos (\gamma + \delta )}\end{array}} \right| = 0$

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

Then, using cos$(A + B) = cosAcosB - sinAsinB$ ,

we get
$\Delta = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{\sin \alpha }&{\cos \alpha }&{\cos \alpha \cos \delta - \sin \alpha \sin \delta }\\{\sin \beta }&{\cos \beta }&{\cos \beta \cos \delta - \sin \beta \sin \delta }\\{\sin \gamma }&{\cos \gamma }&{\cos \gamma \cos \delta - \sin \gamma \sin \delta }\end{array}} \right|$

Applying ${C_3} \to {C_3} + (\sin \delta ){C_1} - (\cos \delta ){C_2},$

we get
$= \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{\sin \alpha }&{\cos \alpha }&0\\{\sin \beta }&{\cos \beta }&0\\{\sin \gamma }&{\cos \gamma }&0\end{array}} \right| = 0$

$(As,\;{C_3} = 0)$

Hence, proved.