Question

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

• A. $2\sin \alpha \sin \beta \sin \gamma$
• B. $4\sin \alpha \sin \beta \sin \gamma$
• C. $\sin \alpha \sin \beta \sin \gamma$
• D. None of these

Answer 1

The correct option is D

None of these

Explanation for the correct option:

Simplifying the equations using trigonometric identities:

$$\begin{gathered} \Rightarrow \sin (\beta +\gamma -\alpha )+\sin (\gamma +\alpha -\beta )+\sin (\alpha +\beta -\gamma )-\sin (\alpha +\beta +\gamma ) \\ \left[\mathrm{Using}\mathrm{formula}e; \\ \sin \left(x-y\right)=\sin x\cos y-\sin y\cos x \\ \sin \left(x+y\right)=\sin x\cos y+\sin y\cos x\right] \\ \Rightarrow \left[\sin (\beta +\gamma )\cos \alpha –\sin \alpha \cos (\beta +\gamma )\right]+\left[\sin (\alpha +\gamma )\cos \beta –\sin \beta \cos (\alpha +\gamma )\right]+ \\ \left[\sin (\alpha +\beta )\cos \gamma –\sin \gamma \cos (\alpha +\beta )\right]–\left[\sin (\alpha +\beta )\cos \gamma +\sin \gamma \cos (\alpha +\beta )\right] \\ \Rightarrow \sin (\beta +\gamma )\cos \alpha –\sin \alpha \cos (\beta +\gamma )+\sin (\alpha +\gamma )\cos \beta –\sin \beta \cos (\alpha +\gamma )+ \\ \sin (\alpha +\beta )\cos \gamma –\sin \gamma \cos (\alpha +\beta )–\sin (\alpha +\beta )\cos \gamma –\sin \gamma \cos (\alpha +\beta ) \\ \Rightarrow [\sin \beta \cos \gamma +\cos \beta \sin \gamma ]\cos \alpha –\sin \alpha [\cos \beta \cos \gamma –\sin \beta \sin \gamma ]+ \\ \cos \beta [\sin \alpha \cos \gamma +\cos \alpha \sin \gamma ]–\sin \beta [\cos \alpha \cos \gamma –\sin \alpha \sin \gamma ]–2\sin \gamma [\cos \alpha \cos \beta –\sin \alpha \sin \beta ] \\ \Rightarrow \cos \alpha \sin \beta \cos \gamma +\cos \alpha \cos \beta \sin \gamma –\sin \alpha \cos \beta \cos \gamma + \\ \sin \alpha \sin \beta \sin \gamma +\sin \alpha \cos \beta \cos \gamma +\cos \alpha \cos \beta \sin \gamma –\cos \alpha \sin \beta \cos \gamma \\ +\sin \alpha \sin \beta \sin \gamma –2\cos \alpha \cos \beta \sin \gamma +2\sin \alpha \sin \beta \sin \gamma \\ \Rightarrow –2\cos \alpha \cos \beta \sin \gamma \end{gathered}$$

Therefore, the correct answer is option (D).