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3.Trigonometrical Ratios, Functions and Identities
medium
$tan\,\, 20^o + tan\,\, 40^o + \sqrt 3\,\, tan\,\, 20^o tan\,\, 40^o$ 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
