If A0, B0 and A+B=frac{pi}{3},then the maxi | vedclass

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(B) Given,$A+B=\frac{\pi}{3}$.Let $y = 	an A 	an B$.Since $B = \frac{\pi}{3} - A$,we have $y = 	an A 	an(\frac{\pi}{3} - A)$.Using the $AM$-$GM$ i

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$\frac{1}{3}$

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(B) Given,$A+B=\frac{\pi}{3}$.Let $y = 	an A 	an B$.Since $B = \frac{\pi}{3} - A$,we have $y = 	an A 	an(\frac{\pi}{3} - A)$.Using the $AM$-$GM$ inequality for positive values $	an A$ and $	an B$:$\frac{	an A + 	an B}{2} \geq \sqrt{	an A 	an B}$.We know that $	an A + 	an B = \frac{\sin(A+B)}{\cos A \cos B} = \frac{\sin(\pi/3)}{\cos A \cos B} = \frac{\sqrt{3}/2}{\cos A \cos B}$.Alternatively,using the identity $	an A 	an B = \frac{\cos(A-B) - \cos(A+B)}{\cos(A-B) + \cos(A+B)}$.For a fixed sum $A+B = \frac{\pi}{3}$,the product $	an A 	an B$ is maximized when $A=B$.Thus,$A = B = \frac{\pi}{6}$.Maximum value $= 	an(\frac{\pi}{6}) 	an(\frac{\pi}{6}) = (\frac{1}{\sqrt{3}})^2 = \frac{1}{3}$.

If $A>0, B>0$ and $A+B=\frac{\pi}{3}$,then the maximum value of $\tan A \tan B$ is

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