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

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

Vedclass · Accepted Answer

$4 - 2\sqrt{3}$

Vedclass · Answer

(B) Given $A + B = \frac{\pi}{6}$. Let $y = 	an A + 	an B$.Using the formula $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we have $	an(\frac{\pi}{6}) = \frac{y}{1 - 	an A 	an B} = \frac{1}{\sqrt{3}}$.Thus,$1 - 	an A 	an B = \sqrt{3}y$,which implies $	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. By the Arithmetic Mean-Geometric Mean $(AM-GM)$ inequality:$\frac{	an A + 	an B}{2} \ge \sqrt{	an A 	an B}$$\frac{y}{2} \ge \sqrt{1 - \sqrt{3}y}$Squaring both sides: $\frac{y^2}{4} \ge 1 - \sqrt{3}y \Rightarrow y^2 + 4\sqrt{3}y - 4 \ge 0$.Solving the quadratic equation $y^2 + 4\sqrt{3}y - 4 = 0$ using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$y = \frac{-4\sqrt{3} \pm \sqrt{48 - 4(1)(-4)}}{2} = \frac{-4\sqrt{3} \pm \sqrt{64}}{2} = -2\sqrt{3} \pm 4$.Since $y > 0$,we take $y \ge 4 - 2\sqrt{3}$.Thus,the minimum value is $4 - 2\sqrt{3}$.

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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