$\cos \alpha \sin (\beta-\gamma)+\cos \beta \sin (\gamma-\alpha)$ $+\cos \gamma \sin (\alpha-\beta)$ is…

$\cos \alpha \sin (\beta-\gamma)+\cos \beta \sin (\gamma-\alpha)$ $+\cos \gamma \sin (\alpha-\beta)$ is equal to
  1. 0
  2. $\frac{1}{2}$
  3. 1
  4. $4 \cos \alpha \cos \beta \cos \gamma$

Solution

$\cos \alpha \sin (\beta-\gamma)+\cos \beta \sin (\gamma-\alpha)$+\cos \gamma \sin (\alpha-\beta)$ $=\cos \alpha[\sin \beta \cos \gamma-\cos \beta \sin \gamma]$+\cos \beta[\sin \gamma \cos \alpha-\cos \gamma \sin \alpha]$+\cos \gamma[\sin \alpha \cos \beta-\cos \alpha \sin \beta]=0$

Asked in: AP EAMCET 2003

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