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Question: \(\tan\frac{2\pi}{5} - \tan\frac{\pi}{15} - \sqrt{3}\tan\frac{2\pi}{5}\tan\frac{\pi}{15}\) is equal ...

tan⁡2π5−tan⁡π15−3tan⁡2π5tan⁡π15\tan\frac{2\pi}{5} - \tan\frac{\pi}{15} - \sqrt{3}\tan\frac{2\pi}{5}\tan\frac{\pi}{15} is equal to

A

−3- \sqrt{3}

B

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

C

1

D

3\sqrt{3}

Answer

3\sqrt{3}

Explanation

Solution

We have tan⁡6π15−tan⁡π151+tan⁡6π15tan⁡π15=tan⁡π3\frac { \tan \frac { 6 \pi } { 15 } - \tan \frac { \pi } { 15 } } { 1 + \tan \frac { 6 \pi } { 15 } \tan \frac { \pi } { 15 } } = \tan \frac { \pi } { 3 }

⇒tan⁡6π15−tan⁡π15=3+3tan⁡6π15tan⁡π15\Rightarrow \tan\frac{6\pi}{15} - \tan\frac{\pi}{15} = \sqrt{3} + \sqrt{3}\tan\frac{6\pi}{15}\tan\frac{\pi}{15}

⇒tan⁡6π15−tan⁡π15−3tan⁡6π15tan⁡π15=3\Rightarrow \tan\frac{6\pi}{15} - \tan\frac{\pi}{15} - \sqrt{3}\tan\frac{6\pi}{15}\tan\frac{\pi}{15} = \sqrt{3}.