If A + B = frac{pi}{4},then (1 + tan A)(1 + t | vedclass

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(B) Given that $A + B = \frac{\pi}{4}$.Taking the tangent on both sides,we get $	an(A + B) = 	an\left(\frac{\pi}{4}ight)$.Using the formula $	an(

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$2$

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(B) Given that $A + B = \frac{\pi}{4}$.Taking the tangent on both sides,we get $	an(A + B) = 	an\left(\frac{\pi}{4}ight)$.Using the formula $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we have $\frac{	an A + 	an B}{1 - 	an A 	an B} = 1$.This implies $	an A + 	an B = 1 - 	an A 	an B$.Rearranging the terms,we get $	an A + 	an B + 	an A 	an B = 1$.Now,consider the expression $(1 + 	an A)(1 + 	an B) = 1 + 	an B + 	an A + 	an A 	an B$.Substituting the value from the previous step,we get $1 + (	an A + 	an B + 	an A 	an B) = 1 + 1 = 2$.

If $A + B = \frac{\pi}{4}$,then $(1 + \tan A)(1 + \tan B) = $

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