MCQ
$\tan \,{20^o}\cot \,{10^o}\cot \,{50^o}$ is equal to
- A$\frac{1}{{\sqrt 3 }}$
- ✓$\sqrt 3 $
- C$\frac{{\sqrt 3 }}{4}$
- D$4\sqrt 3 $
$\tan \theta \tan (60-\theta) \tan (60+\theta)=\tan 3 \theta$
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