Question

If $\alpha ,\beta ,\gamma \in (0,\frac{\pi }{2})$ then the value of $\frac{\sin (\alpha +\beta +\gamma )}{\sin \alpha +\sin \beta +\sin \gamma }$ is

• A. $<1$
• B. $>1$
• C. $=1$
• D. None of these

Answer 1

The correct option is A $<1$

$\sin (\alpha +\beta +\gamma )=\sin \alpha \cos \beta \cos \gamma +\sin \beta \cos \alpha \cos \gamma +\sin \gamma \cos \beta \cos \alpha +\sin \alpha \sin \beta \sin \gamma \therefore \sin \alpha +\sin \beta +\sin \gamma -\sin (\alpha +\beta +\gamma )=\sin \alpha (1-\cos \beta \cos \gamma )+\sin \beta (1-\cos \alpha \cos \gamma )+\sin \gamma (1-\cos \alpha \cos \beta )+\sin \alpha \sin \beta \sin \gamma \ge 0$

As $\alpha ,\beta ,\gamma ϵ(0,\frac{\pi }{2})\Rightarrow 0$

Therefore , $\sin \alpha +\sin \beta +\sin \gamma >\sin \alpha \sin \beta \sin \gamma \Rightarrow \frac{\sin \alpha \sin \beta \sin \gamma }{\sin \alpha +\sin \beta +\sin \gamma }<1$
Hence, option 'A' is correct.