The product (1+tan 1^{circ})(1+tan 2^{circ})( | vedclass

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(C) Let $P = (1+	an 1^{\circ})(1+	an 2^{\circ}) \dots (1+	an 44^{\circ})(1+	an 45^{\circ})$.We use the identity $(1+	an 	heta)(1+	an(45^{\circ}

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$2^{23}$

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(C) Let $P = (1+	an 1^{\circ})(1+	an 2^{\circ}) \dots (1+	an 44^{\circ})(1+	an 45^{\circ})$.We use the identity $(1+	an 	heta)(1+	an(45^{\circ}-	heta)) = 1 + 	an 	heta + 	an(45^{\circ}-	heta) + 	an 	heta 	an(45^{\circ}-	heta)$.Since $	an(45^{\circ}-	heta) = \frac{1-	an 	heta}{1+	an 	heta}$,the expression becomes $1 + 	an 	heta + \frac{1-	an 	heta}{1+	an 	heta} + 	an 	heta \left(\frac{1-	an 	heta}{1+	an 	heta}ight) = 1 + \frac{(1+	an 	heta)^2 + (1-	an 	heta) + 	an 	heta - 	an^2 	heta}{1+	an 	heta} = 1 + \frac{1+2	an 	heta + 	an^2 	heta + 1 - 	an^2 	heta}{1+	an 	heta} = 1 + \frac{2+2	an 	heta}{1+	an 	heta} = 1 + 2 = 3$ (Wait,the identity is $(1+	an 	heta)(1+	an(45^{\circ}-	heta)) = 2$).Pairing terms: $(1+	an 1^{\circ})(1+	an 44^{\circ}) = 2$,$(1+	an 2^{\circ})(1+	an 43^{\circ}) = 2$,...,$(1+	an 22^{\circ})(1+	an 23^{\circ}) = 2$.There are $22$ such pairs,so the product of these is $2^{22}$.Finally,we have the term $(1+	an 45^{\circ}) = 1+1 = 2$.Thus,the total product is $2^{22} 	imes 2 = 2^{23}$.

The product $(1+\tan 1^{\circ})(1+\tan 2^{\circ})(1+\tan 3^{\circ}) \dots (1+\tan 45^{\circ})$ equals

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