Tan ,{20^o}cot ,{10^o}cot ,{50^o} is equal to

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Trigonometrical Equations

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$\tan \,{20^o}\cot \,{10^o}\cot \,{50^o}$ is equal to

A

$\frac{1}{{\sqrt 3 }}$

B

$\sqrt 3 $

C

$\frac{{\sqrt 3 }}{4}$

D

$4\sqrt 3 $

Solution

$\tan 20 . \tan 80 . \tan 40=\tan 3(20)=\sqrt{3} \mathrm{\,\,as}$

$\tan \theta \tan (60-\theta) \tan (60+\theta)=\tan 3 \theta$

Standard 11

Mathematics

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