If (1+tan 1^{circ})(1+tan 2^{circ}) ldots (1+ | vedclass

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(C) Consider $A+B=45^{\circ}$. Then $	an(A+B)=1$,which implies $	an A + 	an B = 1 - 	an A 	an B$. Adding $1 + 	an A 	an B$ to both sides gives

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$23$

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(C) Consider $A+B=45^{\circ}$. Then $	an(A+B)=1$,which implies $	an A + 	an B = 1 - 	an A 	an B$. Adding $1 + 	an A 	an B$ to both sides gives $1 + 	an A + 	an B + 	an A 	an B = 2$,or $(1+	an A)(1+	an B) = 2$. We can pair terms from $1^{\circ}$ to $44^{\circ}$ such that the sum of angles is $45^{\circ}$: $(1+	an 1^{\circ})(1+	an 44^{\circ}) = 2$,$(1+	an 2^{\circ})(1+	an 43^{\circ}) = 2$,...,$(1+	an 22^{\circ})(1+	an 23^{\circ}) = 2$. There are $22$ such pairs,so their product is $2^{22}$. Finally,we include the last term: $(1+	an 45^{\circ}) = 1+1 = 2$. Thus,the total product is $2^{22} 	imes 2 = 2^{23}$. Comparing with $2^n$,we get $n=23$.

If $(1+\tan 1^{\circ})(1+\tan 2^{\circ}) \ldots (1+\tan 45^{\circ})=2^n$,then $n=$

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