3.Trigonometrical Ratios, Functions and Identities
easy

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

A

$< 1$

B

$> 1$

C

$= 1$

D

None of these

Solution

(a) We have $\sin \alpha + \sin \beta + \sin \gamma – \sin (\alpha + \beta + \gamma )$ 

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

$ – \cos \alpha \sin \beta \cos \gamma – \cos \alpha \cos \beta \sin \gamma + \sin \alpha \sin \beta \sin \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 > 0$ 

$\therefore \sin \alpha + \sin \beta + \sin \gamma > \sin (\alpha + \beta + \gamma )$ 

$ \Rightarrow \frac{{\sin (\alpha + \beta + \gamma )}}{{\sin \alpha + \sin \beta + \sin \gamma }} < 1$ .

Standard 11
Mathematics

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