A) \[-\sqrt{3}\]
B) \[\frac{1}{\sqrt{3}}\]
C) 1
D) \[\sqrt{3}\]
Correct Answer: D
Solution :
We have \[\frac{\tan \frac{6\pi }{15}-\tan \frac{\pi }{15}}{1+\tan \frac{6\pi }{15}\tan \frac{\pi }{15}}=\tan \frac{\pi }{3}\] \[\Rightarrow \,\,\tan \frac{6\pi }{15}-\tan \frac{\pi }{15}=\sqrt{3}+\sqrt{3}\,\tan \frac{6\pi }{15}\tan \frac{\pi }{15}\] \[\Rightarrow \,\,\tan \frac{6\pi }{15}-\tan \frac{\pi }{15}-\sqrt{3}\,\tan \frac{6\pi }{15}\tan \frac{\pi }{15}=\sqrt{3}\].
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