The minimum value of frac{tan(x + frac{pi}{6} | vedclass

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(D) Let $y = \frac{	an(x + \frac{\pi}{6})}{	an x}$.Using the formula $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we have:$y = \frac{	an

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$3$

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(D) Let $y = \frac{	an(x + \frac{\pi}{6})}{	an x}$.Using the formula $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we have:$y = \frac{	an x + 	an(\frac{\pi}{6})}{	an x (1 - 	an x 	an(\frac{\pi}{6}))} = \frac{	an x + \frac{1}{\sqrt{3}}}{	an x (1 - \frac{	an x}{\sqrt{3}})} = \frac{\sqrt{3} 	an x + 1}{	an x (\sqrt{3} - 	an x)}$.Alternatively,using sine and cosine:$y = \frac{\sin(x + \frac{\pi}{6}) \cos x}{\cos(x + \frac{\pi}{6}) \sin x} = \frac{2 \sin(x + \frac{\pi}{6}) \cos x}{2 \sin x \cos(x + \frac{\pi}{6})} = \frac{\sin(2x + \frac{\pi}{6}) + \sin(\frac{\pi}{6})}{\sin(2x + \frac{\pi}{6}) - \sin(\frac{\pi}{6})} = \frac{\sin(2x + \frac{\pi}{6}) + 1/2}{\sin(2x + \frac{\pi}{6}) - 1/2}$.Let $u = \sin(2x + \frac{\pi}{6})$. Then $y = \frac{u + 1/2}{u - 1/2} = 1 + \frac{1}{u - 1/2}$.For $y$ to be a local minimum,we need the denominator $u - 1/2$ to be as large as possible while positive,or analyze the function behavior.From the graph,the local minimum value is $3$.

The minimum value of $\frac{\tan(x + \frac{\pi}{6})}{\tan x}$ is

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