If $\tan {3^o} + 2\tan {6^o} + 4\tan {12^o} + 8\cot {24^o} = \cot {\theta ^o}$,then:

  • A
    $\cot (10\theta)^o = 1$
  • B
    $\cot (15\theta)^o = 1$
  • C
    $\cot {\theta ^o} = 0$
  • D
    $\cot {(15\theta)^o} = \sqrt{3}$

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