The value of log tan 1^{circ} + log tan 2^{ci | vedclass

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(B) Using the property of logarithms,$\log a + \log b = \log (ab)$,the expression becomes:$\log (	an 1^{\circ} \cdot 	an 2^{\circ} \cdot 	an 3^{\ci

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(B) Using the property of logarithms,$\log a + \log b = \log (ab)$,the expression becomes:$\log (	an 1^{\circ} \cdot 	an 2^{\circ} \cdot 	an 3^{\circ} \cdot \ldots \cdot 	an 88^{\circ} \cdot 	an 89^{\circ})$We can pair the terms using the identity $	an 	heta \cdot 	an(90^{\circ} - 	heta) = 	an 	heta \cdot \cot 	heta = 1$:$= \log [(	an 1^{\circ} \cdot 	an 89^{\circ}) \cdot (	an 2^{\circ} \cdot 	an 88^{\circ}) \cdot \ldots \cdot (	an 44^{\circ} \cdot 	an 46^{\circ}) \cdot 	an 45^{\circ}]$Since $	an 89^{\circ} = \cot 1^{\circ}$,$	an 88^{\circ} = \cot 2^{\circ}$,and so on:$= \log [(1) \cdot (1) \cdot \ldots \cdot (1) \cdot 	an 45^{\circ}]$Since $	an 45^{\circ} = 1$:$= \log (1 \cdot 1 \cdot \ldots \cdot 1) = \log 1 = 0$

The value of $\log \tan 1^{\circ} + \log \tan 2^{\circ} + \log \tan 3^{\circ} + \ldots + \log \tan 89^{\circ}$ is equal to

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