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Question: \[\tan 20{^\circ} + \tan 40{^\circ} + \sqrt{3}\tan 20{^\circ}\tan 40{^\circ} =\]...

tan⁡20∘+tan⁡40∘+3tan⁡20∘tan⁡40∘=\tan 20{^\circ} + \tan 40{^\circ} + \sqrt{3}\tan 20{^\circ}\tan 40{^\circ} =

A

13\frac{1}{\sqrt{3}}

B

3\sqrt{3}

C

−13- \frac{1}{\sqrt{3}}

D

−3- \sqrt{3}

Answer

3\sqrt{3}

Explanation

Solution

We know that tan⁡(20o+40o)=tan⁡20o+tan⁡40o1−tan⁡20otan⁡40o\tan(20^{o} + 40^{o}) = \frac{\tan 20^{o} + \tan 40^{o}}{1 - \tan 20^{o}\tan 40^{o}}

⇒3=tan⁡20o+tan⁡40o1−tan⁡20otan⁡40o\Rightarrow \sqrt{3} = \frac{\tan 20^{o} + \tan 40^{o}}{1 - \tan 20^{o}\tan 40^{o}}

⇒3−3tan⁡20otan⁡40o=tan⁡20o+tan⁡40o\Rightarrow \sqrt{3} - \sqrt{3}\tan 20^{o}\tan 40^{o} = \tan 20^{o} + \tan 40^{o}

⇒tan⁡20o+tan⁡40o+3tan⁡20otan⁡40o=3.\Rightarrow \tan 20^{o} + \tan 40^{o} + \sqrt{3}\tan 20^{o}\tan 40^{o} = \sqrt{3}.