Tan,, 20^o + tan,, 40^o + √ 3,, tan,, 20^o tan,, 40^o is equal to

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3.Trigonometrical Ratios, Functions and Identities

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$tan\,\, 20^o + tan\,\, 40^o + \sqrt 3\,\, tan\,\, 20^o tan\,\, 40^o$ is equal to

A

$\frac{{\sqrt 3 }}{2}$

B

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

C

$\sqrt 3$

D

$1$

Solution

$ \sqrt{3} =\tan 60^{\circ}=\tan \left(40^{\circ}+20^{\circ}\right) $

$=\frac{\tan 40^{\circ}+\tan 20^{\circ}}{1-\tan 40^{\circ} \tan 20^{\circ}} $

$ \therefore \sqrt{3}-\sqrt{3} \tan 40^{\circ} \tan 20^{\circ}=\tan 40^{\circ}+\tan 20^{\circ} $

Hence $\tan {40^\circ } + \tan {20^\circ } + \sqrt 3 \tan {40^\circ }\tan {20^\circ }$

$=\sqrt{3}$

Standard 11

Mathematics

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