The value of e^{log _{10} tan 1^{circ}+log _{ | vedclass

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(C) Let $S = \log _{10} 	an 1^{\circ} + \log _{10} 	an 2^{\circ} + \ldots + \log _{10} 	an 89^{\circ}$.Using the property $\log a + \log b = \log(a

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(C) Let $S = \log _{10} 	an 1^{\circ} + \log _{10} 	an 2^{\circ} + \ldots + \log _{10} 	an 89^{\circ}$.Using the property $\log a + \log b = \log(ab)$,we have:$S = \log _{10} (	an 1^{\circ} \cdot 	an 2^{\circ} \cdot 	an 3^{\circ} \cdot \ldots \cdot 	an 89^{\circ})$.We know that $	an 	heta \cdot 	an(90^{\circ} - 	heta) = 	an 	heta \cdot \cot 	heta = 1$.Pairing the terms: $(	an 1^{\circ} \cdot 	an 89^{\circ}) = 1, (	an 2^{\circ} \cdot 	an 88^{\circ}) = 1, \ldots, (	an 44^{\circ} \cdot 	an 46^{\circ}) = 1$.The middle term is $	an 45^{\circ} = 1$.Thus,the product is $1 \cdot 1 \cdot \ldots \cdot 1 = 1$.Therefore,$S = \log _{10} (1) = 0$.The expression becomes $e^S = e^0 = 1$.

The value of $e^{\log _{10} \tan 1^{\circ}+\log _{10} \tan 2^{\circ}+\log _{10} \tan 3^{\circ}+\ldots+\log _{10} \tan 89^{\circ}}$ is

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