# Prove trigonometric an identity \frac{2 \cos \alpha-1}{\sqrt 3 -2 \sin \alpha}=\tan(\frac{\alpha}{2}+\frac{\pi}{6})

Prove trigonometric an identity
$\frac{2\mathrm{cos}\alpha -1}{\sqrt{3}-2\mathrm{sin}\alpha }=\mathrm{tan}\left(\frac{\alpha }{2}+\frac{\pi }{6}\right)$
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Joseph Fair
Using Prosthaphaeresis Formulas
$\frac{12}{=}\mathrm{sin}\frac{\pi }{6}=\mathrm{cos}\frac{\pi }{3}$
$\frac{\sqrt{3}}{2}=\mathrm{cos}\frac{\pi }{6}=\mathrm{sin}\frac{\pi }{3}$
$\mathrm{cos}\alpha -\mathrm{cos}\frac{\pi }{3}=2\mathrm{sin}\left(\frac{\pi }{6}-\frac{\alpha }{2}\right)\mathrm{sin}\left(\frac{\pi }{6}+\frac{\alpha }{2}\right)$
$\mathrm{sin}\frac{\pi }{3}-\mathrm{sin}\alpha =2\mathrm{sin}\left(\frac{\pi }{6}-\frac{\alpha }{2}\right)\mathrm{cos}\left(\frac{\pi }{6}+\frac{\alpha }{2}\right)$
Remember we need $\mathrm{sin}\left(\frac{\pi }{6}-\frac{\alpha }{2}\right)\ne 0$ to reach at the identity required
Alternatively, we can use Weierstrass Substitution in the left hand side and use $\mathrm{tan}\left(A+B\right)$ formula
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$\frac{1}{\left(\mathrm{tan}\frac{x}{2}+1{\right)}^{2}{\mathrm{cos}}^{2}\frac{x}{2}}=\frac{1}{\left({\mathrm{tan}}^{2}\frac{x}{2}+2\mathrm{tan}\frac{x}{2}+1\right){\mathrm{cos}}^{2}\frac{x}{2}}\phantom{\rule{0ex}{0ex}}=\frac{1}{{\mathrm{sin}}^{2}\frac{x}{2}+2\mathrm{sin}\frac{x}{2}\mathrm{cos}\frac{x}{2}+{\mathrm{cos}}^{2}\frac{x}{2}}=\frac{1}{1+2\mathrm{sin}\frac{x}{2}\mathrm{cos}\frac{x}{2}}=\frac{1}{1+\mathrm{sin}x}$