Trigonometric Identities Question 359
Question: $ \tan 20{}^\circ +\tan 40{}^\circ +\sqrt{3}\tan 20{}^\circ \tan 40{}^\circ = $
Options:
A) $ \frac{1}{\sqrt{3}} $
B) $ \sqrt{3} $
C) $ -\frac{1}{\sqrt{3}} $
D) $ -\sqrt{3} $
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Answer:
Correct Answer: B
Solution:
We know that $ \tan ,(20^{o}+40^{o})=\frac{\tan ,20^{o}+\tan ,40^{o}}{1-\tan ,20^{o},\tan ,40^{o}} $
$ \Rightarrow ,\sqrt{3}=\frac{\tan ,20^{o}+\tan ,40^{o}}{1-\tan ,20^{o},\tan ,40^{o}} $
$ \Rightarrow \sqrt{3}-\sqrt{3},\tan ,20^{o},\tan ,40^{o}=\tan ,20^{o}+\tan ,40^{o} $
$ \Rightarrow \tan ,20^{o}+\tan ,40^{o}+\sqrt{3},\tan ,20^{o},\tan ,40^{o}=\sqrt{3}. $
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