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  4. tan (2π/5) - tan (π/15) - √3 tan (2π/5) tan (π/15) is equal to

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Q. $\tan \frac{2\pi}{5} -\tan \frac{\pi}{15} - \sqrt{3} \tan \frac{2\pi}{5} \tan \frac{\pi}{15} $ is equal to

Trigonometric Functions

Solution:

Now $ \frac{2\pi}{5} - \frac{\pi}{15} = \frac{\pi}{3} $
$ \Rightarrow \tan \frac{2\pi}{5} - \frac{\pi}{15} =\tan \frac{\pi}{3}$
$ \Rightarrow \frac{\tan \frac{2\pi}{5} -\tan \frac{\pi}{15}}{1+ \tan \frac{2\pi}{5} \tan \frac{\pi}{15}} = \sqrt{3} $