Solveeit Logo
Login

Question

Mathematics Question on Properties of Inverse Trigonometric Functions

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

1

C

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

D

−3 -\sqrt{3}

Answer

3\sqrt{3}

Explanation

Solution

Now 2π5−π15=π3\frac{2\pi}{5} - \frac{\pi}{15} = \frac{\pi}{3} ⇒tan⁡2π5−π15=tan⁡π3 \Rightarrow \tan \frac{2\pi}{5} - \frac{\pi}{15} =\tan \frac{\pi}{3} ⇒tan⁡2π5−tan⁡π151+tan⁡2π5tan⁡π15=3\Rightarrow \frac{\tan \frac{2\pi}{5} -\tan \frac{\pi}{15}}{1+ \tan \frac{2\pi}{5} \tan \frac{\pi}{15}} = \sqrt{3}