If $\sin \alpha ,\sin \beta ,\sin \gamma$ are in A.P. and $\cos \alpha ,\cos \beta ,\cos \gamma$ are in G.P. then $\frac{{\cos }^2\alpha +{\cos }^2\gamma -4\cos \alpha \cos \gamma }{1-\sin \alpha \sin \gamma }=$

From the given conditions we have

$\begin{array}{l}2\sin \beta =\sin \alpha +\sin \gamma \left(i\right)\\ {\cos }^2\beta =\cos \alpha \cos \gamma \left(ii\right)\end{array}$

Squaring (i), $4{\sin }^2\beta ={\sin }^2\alpha +{\sin }^2\gamma +2\sin \alpha \sin \gamma$

Using     (ii), $4(1-\cos \alpha \cos \gamma )=1-{\cos }^2\alpha +1-{\cos }^2\gamma +2\sin \alpha \sin \gamma$

$\begin{array}{l}\Rightarrow {\cos }^2\alpha +{\cos }^2\gamma -4\cos \alpha \cos \gamma =2(\sin \alpha \sin \gamma -1)\\ \Rightarrow \frac{{\cos }^2\alpha +{\cos }^2\gamma -4\cos \alpha \cos \gamma }{1-\sin \alpha \sin \gamma }=-2\end{array}$