Question

If $\sin \alpha ,\sin \beta ,\sin \gamma$ are in AP and $\cos \alpha ,\cos \beta ,\cos \gamma$ are in GP 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

The correct option is A $-2$

$2\times sin\beta =sin\alpha +sin\gamma ,co{s}^2\beta =cos\alpha \times cos\gamma$

By squaring first equation, we get

$4\times {sin}^2\beta ={sin}^2\alpha +{sin}^2\gamma +2\times sin\alpha \times sin\gamma$

Which is equal to $co{s}^2\alpha +{cos}^2\gamma -4\times co{s}^2\beta =-2+2\times sin\alpha \times sin\gamma$

Which implies $co{s}^2\alpha +{cos}^2\gamma -4\times cos\alpha \times cos\gamma =-2\times (1-sin\alpha \times sin\gamma )$

Therefore $\frac{{cos}^2\alpha +{cos}^2\gamma -4\times cos\alpha \times cos\gamma }{(1-sin\alpha \times sin\gamma )}=-2$