$α, β, γ are any 3 angles then \cos α + \cos β - \cos γ - \cos (α + β + γ) =$

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$4 \cos \frac{α + β}{2} \cos \frac{β + γ}{2} \cos \frac{γ + α}{2}$

$4 \cos \frac{α + β}{2} \sin \frac{β + γ}{2} \sin \frac{γ + α}{2}$

$4 \cos \frac{α + β}{2} \cos \frac{β - γ}{2} \cos \frac{γ - α}{2}$

$4 \sin \frac{α + β}{2} \cos \frac{β + γ}{2} \cos \frac{γ - α}{2}$

answer is B.

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Detailed Solution

$L.H.S. = \cos α + \cos β - \cos γ - \cos (α + β + γ) = (\cos α + \cos β) - [\cos γ + \cos (α + β + γ)]$

$\begin{aligned} = 2 \cos (\frac{α + β}{2}) \cos (\frac{α - β}{2}) - 2 \cos (\frac{α + β + γ + γ}{2}) \cos (\frac{α + β + γ - γ}{2}) \\ = 2 \cos (\frac{α + β}{2}) \cos (\frac{α - β}{2}) - 2 \cos (\frac{α + β + γ + γ}{2}) \cos (\frac{α + β + γ - γ}{2}) \\ = 2 \cos (\frac{α + β}{2}) \cos (\frac{α - β}{2}) - 2 \cos (\frac{α + β}{2}) \cos (\frac{α + β + 2 γ}{2}) \end{aligned}$

$= 2 \cos (\frac{α + β}{2}) \{\cos (\frac{α - β}{2}) - \cos (\frac{α + β + 2 γ}{2})\}$

$\begin{aligned} 2 \cos (\frac{α + β}{2}) \{2 \sin (\frac{\frac{α - β}{2} + \frac{α + β + 2 γ}{2}}{2}) \sin (\frac{\frac{α + β + 2 γ}{2} - \frac{α - β}{2}}{2}) \\ = 2 \cos (\frac{α + β}{2}) \{2 \sin (\frac{α + γ}{2}) \sin (\frac{β + γ}{2})\} = \\ = 4 \cos \frac{α + β}{2} \sin \frac{β + γ}{2} \sin \frac{γ + α}{2} = R.H.S. \end{aligned}$