Question

$\cos \alpha \sin (\beta -\gamma )+\cos \beta \sin (\gamma -\alpha )+\cos \gamma \sin (\alpha -\beta )=?$

• A. $0$
• B. $\frac{1}{2}$
• C. $1$
• D. $4\cos \alpha \cos \beta \cos \gamma$

Answer 1

The correct option is A

$0$

Explanation for the correct option:

Solve the expression $\cos \alpha \sin (\beta –\gamma )+\cos \beta \sin (\gamma –\alpha )+\cos \gamma \sin (\alpha –\beta )$

Using the formula $\sin (A-B)=\sin A\cos B-\sin B\cos A$, we get

$=\cos \alpha (\sin \beta \cos \gamma –\cos \beta \sin \gamma )+\cos \beta (\sin \gamma \cos \alpha –\cos \gamma \sin \alpha )+\cos \gamma (\sin \alpha \cos \beta –\cos \alpha \sin \beta )$

$=\cos \alpha \sin \beta \cos \gamma –\cos \alpha \cos \beta \sin \gamma +\cos \alpha \cos \beta \sin \gamma –\sin \alpha \cos \beta \cos \gamma +\sin \alpha \cos \beta \cos \gamma –\cos \alpha \sin \beta \cos \gamma$

$=\mathbf{}\mathbf{0}$

Hence, Option ‘A’ is Correct.