The value of $$4\cos\left(\frac{\pi}{6} - \alpha\right)\sin\left(\frac{\pi}{3} - \alpha\right)$$ is equal to:​

Correct option is D

Given:

​$$4\cos\left(\frac{\pi}{6} - \alpha\right)\sin\left(\frac{\pi}{3} - \alpha\right)$$

Formula used:

​$$\begin{aligned}& \operatorname{Cos}(\theta-\alpha)=(\operatorname{Cos} \theta \operatorname{Cos} \alpha+\operatorname{Sin} \theta \operatorname{Sin} \alpha) \\& \operatorname{Sin}(\theta-\alpha)=(\operatorname{Sin} \theta \operatorname{Cos} \alpha-\operatorname{Cos} \theta \operatorname{Sin} \alpha) \\& (A+B)(A-B)=A^2-B^2 \\& \operatorname{Sin}^2 \theta+\operatorname{Cos}^2 \theta=1 \\& \pi=180^{\circ}\end{aligned}$$  

Solution: 

​$$\begin{aligned}& 4 \operatorname{Cos}\left(\frac{\pi}{6}-\alpha\right) \sin \left(\frac{\pi}{3}-\alpha\right) \\& \Rightarrow 4 \times \operatorname{Cos}\left(30^{\circ}-\alpha\right) \operatorname{Sin}\left(60^{\circ}-\alpha\right) \\& \Rightarrow 4 \times\left(\operatorname{Cos} 30^{\circ} \operatorname{Cos} \alpha+\operatorname{Sin} 30^{\circ} \operatorname{Sin} \alpha\right) \times\left(\operatorname{Sin} 60^{\circ} \operatorname{Cos} \alpha-\operatorname{Cos} 60^{\circ} \operatorname{Sin} \alpha\right) \\& \Rightarrow 4 \times\left(\frac{\sqrt{3}}{2} \operatorname{Cos} \alpha+\frac{1}{2} \operatorname{Sin} \alpha\right) \times\left(\frac{\sqrt{3}}{2} \operatorname{Cos} \alpha-\frac{1}{2} \operatorname{Sin} \alpha\right) \\& \Rightarrow(\sqrt{3} \operatorname{Cos} \alpha+\operatorname{Sin} \alpha) \times(\sqrt{3} \operatorname{Cos} \alpha-\operatorname{Sin} \alpha) \\& \Rightarrow(\sqrt{3} \operatorname{Cos} \alpha)^2-\operatorname{Sin}^2 \alpha \\& \Rightarrow 3 \operatorname{Cos}^2 \alpha-\operatorname{Sin}^2 \alpha \\& \Rightarrow 3 \times\left(1-\operatorname{Sin}^2 \alpha\right)-\operatorname{Sin}^2 \alpha \\& \Rightarrow 3-4 \sin ^2 \alpha\end{aligned}$$