# Prove trigonometric an identity $\frac{2\cos\alpha -1}{\sqrt3-2\sin\alpha}=\tan\left(\frac{\alpha}{2}+\frac{\pi}{6}\right)$

Prove trigonometric an identity
$$\frac{2\cos\alpha -1}{\sqrt3-2\sin\alpha}=\tan\left(\frac{\alpha}{2}+\frac{\pi}{6}\right)$$

My proof:
$$\frac{2\cos\alpha -1}{\sqrt3-2\sin\alpha}=\frac{\cos\alpha-\frac{1}{2}}{\frac{\sqrt3}{2}-\sin\alpha}=\frac{\cos\alpha-\sin\frac{\pi}{6}}{\cos\frac{\pi}{6}-\sin\alpha}$$
I have no idea what to do now

Using Prosthaphaeresis Formulas

$$\dfrac12=\sin\dfrac\pi6=\cos\dfrac\pi3$$

$$\dfrac{\sqrt3}2=\cos\dfrac\pi6=\sin\dfrac\pi3$$

$$\cos\alpha-\cos\dfrac\pi3=2\sin\left(\dfrac\pi6-\dfrac\alpha2\right)\sin\left(\dfrac\pi6+\dfrac\alpha2\right)$$

$$\sin\dfrac\pi3-\sin\alpha=2\sin\left(\dfrac\pi6-\dfrac\alpha2\right)\cos\left(\dfrac\pi6+\dfrac\alpha2\right)$$

Remember we need $$\sin\left(\dfrac\pi6-\dfrac\alpha2\right)\ne0$$ to reach at the identity required

Alternatively, we can use Weierstrass Substitution in the left hand side and use $$\tan(A+B)$$ formula

Use that $$\tan(x+y)=\frac{2\sin(x)\cos(y)}{\cos(x)\cos(y)-\sin(x)\sin(y)}$$