If A 0, B 0 and A + B = frac{pi}{6},then | vedclass

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(A) Given $A + B = \frac{\pi}{6}$. Let $y = 	an A + 	an B$.Using the identity $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we have $\fr

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$4 - 2\sqrt{3}$

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(A) Given $A + B = \frac{\pi}{6}$. Let $y = 	an A + 	an B$.Using the identity $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we have $\frac{1}{\sqrt{3}} = \frac{y}{1 - 	an A 	an B}$.Thus,$	an A 	an B = 1 - \sqrt{3}y$.Since $A, B > 0$ and $A+B = \frac{\pi}{6}$,both $	an A$ and $	an B$ are positive,so $	an A 	an B > 0$,which implies $1 - \sqrt{3}y > 0$,or $y By $AM \ge GM$,we have $\frac{	an A + 	an B}{2} \ge \sqrt{	an A 	an B}$.Substituting $y$,we get $\frac{y}{2} \ge \sqrt{1 - \sqrt{3}y}$.Squaring both sides,$\frac{y^2}{4} \ge 1 - \sqrt{3}y$,so $y^2 + 4\sqrt{3}y - 4 \ge 0$.The roots of $y^2 + 4\sqrt{3}y - 4 = 0$ are $y = \frac{-4\sqrt{3} \pm \sqrt{48 + 16}}{2} = -2\sqrt{3} \pm 4$.Since $y > 0$,we must have $y \ge 4 - 2\sqrt{3}$.

If $A > 0, B > 0$ and $A + B = \frac{\pi}{6}$,then the minimum value of $\tan A + \tan B$ is

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