$\tan \,{20^o}\cot \,{10^o}\cot \,{50^o}$ is equal to
$\tan \,{20^o}\cot \,{10^o}\cot \,{50^o}$ is equal to
- A
$\frac{1}{{\sqrt 3 }}$
- B
$\sqrt 3 $
- C
$\frac{{\sqrt 3 }}{4}$
- D
$4\sqrt 3 $
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