Evaluate: $log\ sin1^o.log\ sin 2^o. log\ sin3^o\ ........log\ sin90^o$.

Given: $log\ sin1^o.log\ sin 2^o. log\ sin3^o\ ........log\ sin90^o$.

To do: To evaluate: $log\ sin1^o.log\ sin 2^o. log\ sin3^o\ ........log\ sin90^o$.

Solution: 

$log\ sin1^o.log\ sin 2^o. log\ sin3^o\ ........log\ sin90^o$.

$=log\ sin1^o.log\ sin 2^o. log\ sin3^o\ ........log\ ( 1)$           [$\because sin90^o=1$]

$=log\ sin1^o.log\ sin 2^o. log\ sin3^o\ .......\times 0$      [$\because\ log(1)=0$]

$=0$  
 

Thus, $log\ sin1^o.log\ sin 2^o. log\ sin3^o\ ........log\ sin90^o=0$.

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Updated on: 10-Oct-2022

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