Q. $\tan{\frac{2\pi}{5} - \tan{\frac{\pi}{15} - \sqrt{3}}\tan{\frac{2\pi}{5}\tan\frac{\pi}{15}}}$ is equal to
A) $- \sqrt{3}$
B) $\frac{1}{\sqrt{3}}$
C) 1
D) $\sqrt{3}$
Explanation
(d)
Now,${\ tan}{\frac{\pi}{3} = \tan{\left( \frac{6\pi}{15} - \frac{\pi}{15} \right) = \frac{\tan{\frac{6\pi}{15} - \tan\frac{\pi}{15}}}{1 + \tan{\frac{6\pi}{15}\tan\frac{\pi}{15}}}}}$
$\Rightarrow \tan{\frac{6\pi}{15} - \tan{\frac{\pi}{15} = \sqrt{3} + \sqrt{3}\tan{\frac{6\pi}{15}\tan\frac{\pi}{15}}}}$
$\Rightarrow \tan{\frac{6\pi}{15} - \tan{\frac{\pi}{15} - \sqrt{3}\tan{\frac{6\pi}{15}\tan{\frac{\pi}{15} = \sqrt{3}}}}}$
$= \tan{\frac{2\pi}{5} - \tan{\frac{\pi}{15} - \sqrt{3}\tan{\frac{2\pi}{5}\tan{\frac{\pi}{15} = \sqrt{3}}}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.