MCQ
$\cos \alpha .\sin (\beta - \gamma ) + \cos \beta .\sin (\gamma - \alpha ) + \cos \gamma .\sin (\alpha - \beta ) = $
  • $0$
  • B
    $1/2$
  • C
    $1$
  • D
    $4\cos \alpha \cos \beta \cos \gamma $

Answer

Correct option: A.
$0$
a
(a) $\cos \alpha \sin (\beta - \gamma ) + \cos \alpha \sin (\gamma - \alpha ) + \cos \gamma \sin (\alpha - \beta )$

Put $\alpha = \beta = \gamma = {60^o} $

$\Rightarrow \frac{1}{2}(0) + \frac{1}{2}(0) + \frac{1}{2}(0) = 0$.

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