sumlimits_{r = 1}^{89} {{log _3}(tan {r^o})} | vedclass

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(D) Let $S = \sum\limits_{r = 1}^{89} {{\log _3}(	an {r^o})}$.Using the property of logarithms,$\sum \log a = \log (\prod a)$,we get:$S = {\log _3}(\

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(D) Let $S = \sum\limits_{r = 1}^{89} {{\log _3}(	an {r^o})}$.Using the property of logarithms,$\sum \log a = \log (\prod a)$,we get:$S = {\log _3}(	an {1^o} \cdot 	an {2^o} \cdot 	an {3^o} \cdots 	an {89^o})$.We know that $	an {r^o} \cdot 	an {(90 - r)^o} = 1$.Pairing the terms from the product: $(	an {1^o} \cdot 	an {89^o}) \cdot (	an {2^o} \cdot 	an {88^o}) \cdots (	an {44^o} \cdot 	an {46^o}) \cdot 	an {45^o}$.Since each pair equals $1$ and $	an {45^o} = 1$,the product is $1 \cdot 1 \cdots 1 = 1$.Therefore,$S = {\log _3}(1) = 0$.

$\sum\limits_{r = 1}^{89} {{\log _3}(\tan {r^o})} = $

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