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$\log \tan 1^o + \log \tan 2^o + \dots + \log \tan 89^o$ is $\ldots$

  1. $1$
  2. $1/\sqrt{2}$
  3. $0$
  4. $−1$
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$(\tan 1 . \tan 89) . (\tan 2 . \tan 88) \ldots (\tan 44 . \tan 46) . \tan 45$
$= ( \tan 1 . \cot 1).(\tan 2 . \cot 2)\ldots (\tan 44. \cot 44). \tan 45$
$= 1 . 1 . 1 \ldots . 1 . \tan 45$
$= 1$
So, the given expression reduces to $\log 1 = 0.$

Correct Option: (C) 0
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