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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

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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

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

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