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

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

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$2$

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(B) Given that $A + B = \frac{\pi}{4}.$Taking 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 $	an A + 	an B + 	an A 	an B = 1$ into the expression,we get $1 + 1 = 2.$Thus,$(1 + 	an A)(1 + 	an B) = 2.$

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

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