$\tan {3^o} + 2\tan {6^o} + 4\tan {12^o} + 8\cot {24^o} = \cot {\theta ^o}$ then
$\tan {3^o} + 2\tan {6^o} + 4\tan {12^o} + 8\cot {24^o} = \cot {\theta ^o}$ then
- A
$cot (10\ \theta )^o =1$
- B
$cot (15\ \theta )^o =1$
- C
$\cot {\theta ^o} = 0$
- D
$\cot {\left( {15\ \theta } \right)^o} = \sqrt 3$
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