Question

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

• A. $<1$
• B. $=-1$
• C. $<0$
• D. None of these

Answer 1

The correct option is A

$<1$

Explanation for the correct option:

Find the value of $\frac{\sin (\alpha +\beta +\gamma )}{\sin \alpha +\sin \beta +\sin \gamma }$:

As we know,

$\begin{array}{rcl}\sin \alpha +\sin \beta +\sin \gamma -\sin (\alpha +\beta +\gamma )& =& \sin \alpha +\sin \beta +\sin \gamma -[\sin \alpha \cos (\beta +\gamma )+\cos \alpha \sin (\beta +\gamma \left)\right]\\ & =& \sin \alpha +\sin \beta +\sin \gamma –\sin \alpha [\cos \beta \cos \gamma –\sin \beta \sin \gamma ]–\cos \alpha [\sin \beta \cos \gamma +\cos \beta \sin \gamma ]\\ & =& \sin \alpha (1-\cos \beta \cos \gamma )+\sin \beta (1-\cos \gamma \cos \alpha )+\sin \gamma (1-\cos \alpha \cos \beta )+\sin \alpha \sin \beta \sin \gamma >0\end{array}$

$\Rightarrow$$\sin \alpha +\sin \beta +\sin \gamma >\sin (\alpha +\beta +\gamma )$ $\left[\mathbf{\because }\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{\alpha }\mathbf{,}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{\beta }\mathbf{,}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{\gamma }\mathbf{,}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{\alpha }\mathbf{,}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{\beta }\mathbf{,}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{\gamma }\mathbf{}\mathbf{\in }\mathbf{}\left[0,1\right]\right]$

$\mathbf{\therefore }\frac{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{(}\mathbf{\alpha }\mathbf{}\mathbf{+}\mathbf{}\mathbf{\beta }\mathbf{}\mathbf{+}\mathbf{}\mathbf{\gamma }\mathbf{)}}{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{\alpha }\mathbf{}\mathbf{+}\mathbf{}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{\beta }\mathbf{}\mathbf{+}\mathbf{}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{\gamma }}\mathbf{}\mathbf{<}\mathbf{}\mathbf{1}$

Hence, Option ‘A’ is Correct.