Question

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

• A. $-2$
• B. $-1$
• C. $0$
• D. $2$

Answer 1

Answer: (A)

$-2$

$\sin \alpha ,\sin \beta ,\sin \gamma$ are in AP

$2\sin \beta =(\sin \alpha +\sin \gamma )$

$\cos \alpha ,\cos \beta ,\cos \gamma$ are in GP

${\cos }^2\beta =\left(\cos \alpha \cos \gamma \right)$

$4{\sin }^2\beta ={\sin }^2\alpha +{\sin }^2\gamma +2\sin \alpha \sin \gamma \frac{({\cos }^2\alpha +{\cos }^2\gamma -2)+4{\sin }^2\beta }{2}=\left(\sin \alpha \sin \gamma \right)\frac{{\cos }^2\alpha +{\cos }^2\gamma -4\cos \alpha \cos \gamma }{1-\frac{\left[\right({\cos }^2\alpha +{\cos }^2\gamma -2)+4{\sin }^2\beta ]}{2}}=2\left[\frac{{\cos }^2\alpha +{\cos }^2\gamma -4(1-{\sin }^2\beta )}{2-({\cos }^2\alpha +{\cos }^2\gamma -2)+4{\sin }^2\beta }\right]=2\left[\frac{{\cos }^2\alpha +{\cos }^2\gamma +4{\sin }^2\beta -4}{4-({\cos }^2\alpha +{\cos }^2\gamma -2)+4{\sin }^2\beta }\right]=-2$