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Mathematics Question on Trigonometric Functions

If α,β,γ\alpha, \beta, \gamma are any three angles, then cos⁡ α+cos⁡β−cos⁡ γ−cos⁡(α+β+γ)=\cos \, \alpha + \cos \beta - \cos \, \gamma - \cos (\alpha + \beta + \gamma) =

A

4cos⁡α+β2cos⁡β+γ2cos⁡γ+α24 \cos \frac{\alpha+\beta}{2} \cos \frac{\beta+\gamma}{2} \cos \frac{\gamma+\alpha}{2}

B

4cos⁡α+β2sin⁡β+γ2sin⁡γ+α24 \cos \frac{\alpha+\beta}{2} \sin\frac{\beta+\gamma}{2} \sin\frac{\gamma+\alpha}{2}

C

4cos⁡α+β2sin⁡β−γ2sin⁡γ−α24 \cos \frac{\alpha+\beta}{2} \sin\frac{\beta - \gamma}{2} \sin\frac{\gamma - \alpha}{2}

D

4sin⁡α+β2cos⁡β+γ2cos⁡γ+α24 \sin \frac{\alpha+\beta}{2} \cos \frac{\beta+\gamma}{2} \cos \frac{\gamma+\alpha}{2}

Answer

4cos⁡α+β2sin⁡β+γ2sin⁡γ+α24 \cos \frac{\alpha+\beta}{2} \sin\frac{\beta+\gamma}{2} \sin\frac{\gamma+\alpha}{2}

Explanation

Solution

Answer (b) 4cos⁡α+β2sin⁡β+γ2sin⁡γ+α24 \cos \frac{\alpha+\beta}{2} \sin\frac{\beta+\gamma}{2} \sin\frac{\gamma+\alpha}{2}