Question

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

• A. $-2$
• B. $-1$
• C. $0$
• D. $2$

Answer 1

The correct option is D $2$

Assume $y=\frac{{\cos }^2\alpha +{\cos }^2\gamma -4\cos \alpha .\cos \gamma }{1-\sin \alpha .\sin \gamma }$
since, $\sin \alpha ,\sin \beta ,\sin \gamma$ are in A.P $\Rightarrow 2\sin \beta =\sin \alpha +\sin \gamma$
squaring, $4{\sin }^2\beta ={\sin }^2\alpha +{\sin }^2\gamma +2\sin \alpha .\sin \gamma =2-({\cos }^2\alpha +{\cos }^2\gamma )+2\sin \alpha .\sin \gamma$
and $\cos \alpha ,\cos \beta ,\cos \gamma$ are in G.P $\Rightarrow {\cos }^2\beta =\cos \alpha .\cos \gamma$
Using this equation,
$y=\frac{{\cos }^2\alpha +{\cos }^2\gamma -4\cos \alpha .\cos \gamma }{1+\sin \alpha .\sin \gamma }$
$=\frac{2+2\sin \alpha .\sin \gamma -4{\sin }^2\beta -4{\cos }^2\beta }{1+\sin \alpha .\sin \gamma }$
$=\frac{2+2\sin \alpha .\sin \gamma }{1+\sin \alpha .\sin \gamma }=2$